THE BIBLICAL TIMELINE

THE ALGEBRA COMPLETED

Here we finish setting up our algebra for the Hebrew Kings begun on October 29:

 

If n^• is the total number of years a king reigned then we define the year his reign ended when counting from p as P = p(n^•); similarly we define the year his reign ended when counting from q as, Q = q(n^•). When re- ferring to the kings of Israel we use P_I = p_I(n^•) or Q_I. = q_I(n^•).

 

1Q_I or 1P_I stands for the end of the reign of the first king of Israel after Solomon, i.e., after the breakup of the nation of Israel. 2Q_I or 2P_I refers to the end of the reign of the second king of Israel after Solomon etc. Similarly 2P or 2Q refer to the end of the reign of the second king of Judah etc.

 

Note that if Q = q(n^•), then P ≠ p(n^•), rather P = p(n^• + 1). Similarly if P = p(n^•), then Q ≠ q(n^•), rather Q = (n^• – 1). Therefore if we're given n^• years as the total number of years a king reigned, then either Q = q(n^•) is true, or P = p(n^•) is true, but they can't both be true; we can't have a king's reign ending in two different years.

 

Whenever a Biblical verse tells us a king's reign ended in n^• years when counting from q, we always use the symbol, Q, or Q = q(n^•) in respect to that king, not P or P = p(n^• + 1). Although the arithmetic alone is correct in Q = q(n^•) = P = p(n^• + 1), using P is incorrect for the obvious reason that the king reigned for n^•—not (n^•+ 1)—years. Q = q(n^•) tells us we're counting from q, not p, in respect to a certain king, and that's what we want to remember. Similarly whenever a Biblical verse tells us a king's reign ended in n^• years when counting from p, we always use the symbol, P = p(n^•) in respect to that king, not Q = q(n^• – 1), because again, n^• is the significant number, not (n^•– 1).

 

Generally if a, b, are statements then the statement a or b—i.e., (a ∆ b)— could just as well be written as b or a—i.e., (b ∆ a). Another way of saying the same thing is to write, a ∆ b = b ∆ a. Therefore it wouldn't violate our arithmetic or logic to write p(n^•) ∆ q(n^•) = q(n^•) ∆ p(n^•). However to avoid confusion with our numbers we always consider p(n^•) and q(n^•) together as an ordered pair such that p(n^•) always comes first and q(n^•) comes second. To show that p(n^•) & q(n^•) are well-ordered, we put them in parentheses: (p(n^•) ∆ q(n^•)). Therefore we will never write

(q(n•) ∆ p(n•)). Similarly we always write (P ∆ Q), never (Q ∆ P). The same well-ordering applies when using &.

 

We can use simple logic to determine the reigns of the Hebrew kings be- cause a king either reigned in a certain year or he didn't; i.e., we have only two truth values: true or false.

 

The symbol ∆ can be read as, or.

Because (p ∆ q) is true if and only if either p is true or q is true, or both are true, and because Rehoboam's reign ended in either in 1P or 1Q, for some n^•, ((1p(n^•) ∆ 1q(n^•) = (1P ∆ 1Q) is true.

 

We use the tilde, ~, to change the truth value of a statement; and ~ can be read as, "it's not the case that." The symbol & can be read as, and.

(p & q) is true if and only if both p is true and q is true. Because Rehoboam's reign could not end in two different years,

(1P = 1p(n^•) & 1Q = 1q(n^•)) must be false; therefore

 ~(1P = 1p(n^•) & 1Q = 1q(n^•)) is true

 

The symbol → can be read as, implies; or, which implies. The only way the statement p → q can be false is for p to be true and q to be false. Otherwise the statement p → q is true. Because we can't use both P and Q when referring to the end of a king's reign, P → Q is false and Q → P is false. Therefore ~(P→Q) & (Q→P)) is true and (~(P→Q) & ~(Q→P)) is true and (P →~Q & Q →~P) is true.

The symbol iff or ↔ can be read as, if and only if; i.e., if p↔q is true then p→q is true and q→p is true (and conversely).

 

Looking again at 2 Kings 8:25 and 2 kings 9:29:

²⁵ In the twelfth year of Joram the son of Ahab king of Israel did Ahaziah the son of Jehoram king of Judah begin to reign (2 Kings 8).

²⁹And in the eleventh year of Joram the son of Ahab began Ahaziah to reign over Judah (2 Kings 9).

 

Ahaziah was the 6^th king of Judah, i.e., Ahaziah≡6p; and Joram was the 10^th king of Israel, i.e., Joram≡10p_I. If we remember that p(n + 1) = q(n) we can put these two verses into our language and get:

(2 Kings 8:25 & 2 Kings 9:29) → 6p = 10p_I (12) = 10q_I (11).

 

Once we figure out when these kings began to reign, we can plug the numbers into the above equations to prove equality. Forty reigns and their co-regencies offer many possibilities for a contradiction. While one such error would put us out of business, finding none would prove nothing. But at least showing harmony in every possible permutation proves we can determine the Hebrews kings from the Bible alone; and that should boost Scripture up a notch on the credibility scale, as well as put our atheist friends back to sleep.

 

On November 3 we begin with the first kings of Judah and Israel: Rehoboam and Jerobaom.

THE REIGNS OF THE HEBREW KINGS

Rehoboam Jerobaom Abijah

We now use the algebra and logic developed on October 29 and Novem- ber 1 to begin calculating the Hebrew kings. Solomon, the last king to rule before the breakup of national Israel, died in 10082 and we begin there. We have the first king of Judah, 1p, and the first king of Israel, 1p_I, begin- ning at the same time; i.e., 1p = 1p_I = 10082 (see October 22 & 25).

 

(The explanations in green serve not to validate or strengthen the argu- ments but to aid those lacking the equivalent of a course in high school algebra and/or plane geometry. The reasoning and proofs stand on the yellow and aqua alone and most readers probably won't need to consult the green (or fuchsia) at all.)

 

1 Kings 11:³¹And he said to Jeroboam…thus saith the LORD…I will rend the kingdom out of the hand of Solomon, and will give ten tribes to thee:

→ Jeroboam≡1p_I = Solomon = (P ∆ Q) = 10082.

We use 1p_I rather than 1q_I because Jeroboam would have sat on the throne the same year Solomon died, not the following year, 1q_I. We don't need to distinguish between P and Q in the case of Solomon.

 

1 Kings 11:⁴³And Solomon slept with his fathers, and was buried in the city of David his father: and Rehoboam his son reigned in his stead. →

Rehoboam≡1p = Solomon = (P ∆ Q) = 10082.

 

1 Kings 14:²⁰And the days which Jeroboam reigned were two and twenty years: and he slept with his fathers, and Nadab his son reigned in his stead.

→ (1p_I(22) ∆ 1q_I(22)) = (1P_I ∆ 1Q_I) ≥ 2p_I≡Nadab.

This says that either Jeroboam's reign ended in 22 years when counting from point 1p_I is true, or Jeroboam's reign ended in 22 years when count- ing from point 1q_I is true. We use ≥ (meaning greater than or equal to) be- cause we don't know yet whether Nadab reigned as a co-regent with Jero- boam or began with Jeroboam's death.

 

1 Kings 14:²¹…Rehoboam the son of Solomon…was forty and one years old when he began to reign, and he reigned seventeen years in Jerusalem… →

(1p(17) ∆ 1q(17) = (1P ∆ 1Q).

Again this says that either Rehoboam's reign ended in 17 years when counting from 1p is true or Rehoboam's reign ended in 17 years when counting from point 1q is true.

In other words, 1p(17) = 1P is true or 1q(17) = 1Q is true.

(Our definitions of 1Q and 1P automatically rule out such considerations as, 1p(17) = 1Q and 1q(17) = 1P)

 

In 2 Chronicles 12:16: …Rehoboam slept with his fathers, and was buried in the city of David: and Abijah his son reigned in his stead & (1 Kings 14:21) →

Abijah≡2p ≤ (1p(17) ∆ 1q(17) = (1P ∆ 1Q)

Note we need both verses for our conclusion. Again we write Abijah≡2p ≤ (1P ∆ 1Q) because we don't know yet if Abijah reigned as a co-regent with Rehoboam before the death of the latter.

 

1 Kings 15:¹Now in the eighteenth year of king Jeroboam…reigned Abijam (= Abijah) over Judah. ²Three years reigned he in Jerusalem…→

2p = (1p_I(18) ∆ 1q_I(18)) & (2p(3) ∆ 2q(3)) = (2P ∆ 2Q)

This says that either 2p = 1p_I(18) is true or 2p = 1q_I(18) is true and either

2p(3) = 2P is true or 2q(3) = 2Q is true.

 

1)  Suppose 2p = 1q_I(18).

Here we begin a 5 step proof by contradiction:

we suspect that 2p = 1p_I(18) is true, but for the sake of argument we as- sume for the moment that 2p = 1p_I(18) is false. Then 2p would have to equal 1q_I(18). We then show that 2p = 1q_I(18) can't be true because it leads to a contradiction (that Divinity won't allow). That forces us to con- clude that 2p = 1p_I(18) is true.

 

2)  1q_I = 1q → 1q_I (18) = 1q(18)

Here we use one of our corollaries from October 29:

      q_I(n^*) = q(n^*) ↔ q_I = q.

 

3)  2p = 1q(18)

Since we supposed in 1) that 2p = 1q_I(18), and we've just shown in 2) that

1q_I (18) = 1q(18), we can write that 2p = 1q(18).

 

4)  But 2p = 1q(18) contradicts

    (1 Kings 14:21 & 2 Chronicles 12:16) → 2p ≤ (1p(17) ∆ 1q(17)).

Here we have our contradiction: 2p can't be equal to 1q(18) and still be less than or equal to either 1p(17) or 1q(17).

 

5)  ~(2p = 1q_I(18)) → (2p = 1p_I(18))

We can now conclude that our assumption in 1), that 2p = 1q_I(18), is false. Therefore from (1 Kings 15:1) → 2p = (1p_I(18) ∆ 1q_I(18)), we now know that 2p = 1p_I(18) is true.

 

6)   (2p = 1p_I(18)) = ((10082 – 1) + 18) = 10099.

Recall from October 29 that when counting from p we use the formula

p(n) = ((x – 1) + n).

 

7)  1p_I (18) = 1p(18) = 1q(17) = 2p = 10099 and 1q(17) = 1Q.

To get 1p_I (18)) = 1p(18) we use the corollary p_I(n) = p(n)) ↔ (p_I = p)).

To get 1p(18) = 1q(17) we use the corollary p(n) = q(n – 1)

We already know that 1p_I (18) = 2p from step 5).

We already know that 1p_I(18) = 10099 from step 6)

We get 1q(17) = 1Q because 1 Kings 14:21 gives the first king of Judah reigning 17 years. Note that the and is necessary in 7). We couldn't re- place the and 1q(17) with just = . Although the arithmetic may be cor- rect in 1p(18) = 1Q, the first king reigned for 17, not 18 years; further- more 1p(18) = 1Q violates our definition of 1Q; i.e., we're counting from q, not p.

 

8)   1 Kings 15:1 → 2q(3) = (10099 + 3) = 10102 = 2Q

Since we're counting from point 1q we use the formula q(n) = x + n.

 

9)   1 Kings 14:20 → 1p_I (22) = ((10082 – 1) + 22) = 10103 = 1P_I →

       2p_I ≤ 10103

Step 5) gives us 2p = 1p_I(18); therefore we're counting from point 1p_I in respect to the kings of Israel.

1 Kings 14:20 tells us Jeroboam reigned for 22 years.

Therefore 1p_I (22) = 1P_I

1 Kings 14:20 also gives us 2p_I ≤ 1P_I

 

We now have:

1q≡Rehoboam reigned for 17 years: from 10082 until 10099 (= 1Q).

1p_I≡Jerobaom reigned for 22 years: from 10082 until 10103 (= 1P_I).

2q≡Abijah reigned for 3 years: from 10099 until 10102 (= 2Q).

 

On November 5 we look at the kings: Asa, Nadab and Baash

ASA NADAB and BAASH

In continuing the Hebrew kings, we need to know the following (from November 3):

1q≡Rehoboam reigned for 17 years: from 10082 until 10099 (= 1Q).

1p_I≡Jeroboam reigned for 22 years: from 10082 until 10103 (= 1P_I).

2q≡Abijah reigned for 3 years: from 10099 until 10102 (= 2Q).

 

(That in green is for those needing help in understanding the formal argu- ments in aqua.)

 

1 Kings 15: ⁸And Abijam (= Abijah) slept with his fathers; and they buried him in the city of David: and Asa his son reigned in his stead →

Abijah≡2Q = 10102 ≥ 3p≡Asa.

 

1 Kings 15: ⁹…in the twentieth year of Jeroboam king of Israel reigned Asa over Judah →

3p = (1p_I(20) ∆ 1q_I(20))

3p = ((10082 – 1) + 20) ∆ (10082 + 20) →

3p = (10101 ∆ 10102) →

Since 1 Kings 15:8 → 3p ≤ 10102

1 Kings 15:8 agrees with 1 Kings 15:9

 

We've shown that 1 Kings 15:9 implies that either 3p = 10101 is true or 3p = 10102 is true. Since 1 Kings 15:8 implies 3p ≤ 10102, we've shown that 1 Kings 15:8 and 1 Kings 15:9 don't conflict, therefore 1 Kings 15:8 and 15:9 harmonize. Proving this harmony is admittedly a small step; but it's a necessary one in showing that no numerical conflict exists in Scrip- ture regarding the Hebrew kings.

 

1 Kings 15: ¹⁰ And forty and one years reigned he (Asa) in Jerusalem….

Suppose 3p = 2q(3) = 2Q = 10102

 

Abijah≡2q(3) = 2Q reigned for only 3 years. In order to proceed we sup- pose the next king, Asa≡3p, didn't have time to reign as a co-regent with Abijah≡2q(3) = 2Q and consequently Asa≡3p sat on the throne the same year Abijah≡2q(3) = 2Q died. We hope that somewhere down the road our arithmetic will show that Asa≡3p = 10102 is indeed true (or false).

 

1 Kings 15:10 → 3q(41) = 3Q = (10102 + 41) = 10143.

3q≡Asa reigned for 41 years: from 10102 until 10143 (= 3Q).

 

We know from November 3 (see step 7) that we're counting from q in respect to the kings of Judah. Therefore when 1 Kings 15:10 gives us 41 years we use the formula 3q(n) = x + n (see October 29).

 

1 Kings 15: ²⁵And Nadab the son of Jeroboam began to reign over Israel in the second year of Asa king of Judah, and reigned over Israel two years. →

2p_I ≤ 10103 (= 1P_I) = Asa≡(3p(2) ∆ 3q(2)).

2p_I ≤ 10103 = ((10102 – 1) + 2) ∆ (10102 + 2) →

2p_I ≤ 10103 = (10103 ∆ 10104) →

2p_I = 10103 = 3p(2)

2p_I(2) = ((10103 – 1) + 2) = 10104.

2p_I≡Nadab reigned for 2 years: from 10103 until 10104 (= 2P)

 

From November 3 the first line of aqua has Jeroboam≡1P_I = 10103; and 1 Kings 15:25 tells us that Nadab replaced Jeroboam≡1P_I. So we have Nadab≡2p_I ≤ 1P_I. 1 Kings 15:25 also tells us Nadab≡2p_I began to rule in the second year of Asa≡3p. But we don't know if 1 Kings 15:25 refers to the second year of Asa≡3p when counting from point p or when counting from point q. So we write Nadab≡2p_I = Asa≡(3p(2) ∆ 3q(2)). Since we know that 2p_I ≤ 10103, we know that 2p_I ≠ 10104 in the 3^rd line of aqua; i.e., 2p_I≡Nadab wouldn't have begun to rule a year after Jeoboam stepped down; so Nadab≡2p_I could not have begun to reign in the second year of Asa when counting from point q. That leaves us only with Asa≡3p(2) = 2p_I = 10103. 1 Kings 15:25 also tells us that Nadab≡2p_I ruled for two years over Israel. We know from November 3 that we're counting from P_I in respect to the kings of Israel; therefore in the sixth line of aqua we know Nadab≡2p_I(2) = 2 P_I = 10104.

 

1 Kings 15: ²⁸Even in the third year of Asa king of Judah did Baasha slay him (Nadab (1 Kings 15:27)), and reigned in his stead. →

Nadab≡2P_I = 10104 = (3p(3) ∆ 3q(3))≡Asa.

2P_I = 10104 = ((10102 – 1) + 3) ∆ (10102 + 3)) →

2P_I = 10104 = (10104 ∆ 10105)

2P_I = 3p(3) = 10104

1 Kings 15:25 agrees with 1 Kings 15:28; i.e., 2p_I(2) = 2P_I = 10104.

Baasha≡3p_I = 10104.

 

Baasha≡3p_I began to rule in the year Nadab≡2P_I because 1 Kings 15:28 tells us Baasha killed Nadab.

 

1 Kings 15: ³³In the third year of Asa king of Judah began Baasha the son of Ahijah to reign over all Israel in Tirzah, twenty and four years. & (results of 1 Kings 15:28)→

Baasha≡3p_I = 10104 = 3p(3) = 10104. 1 Kings 15:25, 1 Kings 15:28 and 1 Kings 15:33 all agree.

Baasha≡3p_I(24) = ((10104 – 1) + 24) = 10127

3p_I≡Baash reigned for 24 years: from 10104 until 10127 (= 3P_I).

 

2 Chronicles 16: ¹In the six and thirtieth year of the reign of Asa Baasha king of Israel came up against Judah, and built Ramah, to the intent that he might let none go out or come in to Asa king of Judah

3q≡Asa reigned for 41 years: from 10102 until 10143 (= 3Q).

(3p(36) ∆ 3q(36) =) ((10102 – 1) + 36) ∆ 10102 + 36) =

(10137 ∆ 10138)

But the years (10137 ∆ 10138) contradict 3P_I = 10127.

 

Because Baasha didn't worship Jehovah, many Israelites defected to Ju- dah. Therefore Baasha set up a blockade. One thing led to another until Asa and Baasha were at each other's throats.

 

But here we have an apparent contradiction. Baasha ruled until only 10127 (= 3P_I). But the 36^th year of Asa is 10137 or 10138, when Baasha was no longer around. Therefore to find harmony between 2 Chronicles 16:1 and 1 Kings 15:10 we'd have to begin counting at least 10 years be- fore Asa ever began to rule; that would make year 36 during the reign of Asa land on 10127 instead of 10137. So in order to harmonize 2 Chron- icles 16:1 with the other verses, we need to search for the year, p^•, in which to begin counting.

 

We know Asa and Baasha fought each other: ³²…there was war between Asa and Baasha king of Israel all their days (1 Kings 15).

However when ¹…Abijah slept with his fathers…Asa his son reigned in his stead. In his days the land was quiet ten years ( 2 Chronicles 14).

¹⁹And there was no more war unto the five and thirtieth year of the reign of Asa (2 Chronicles 15).

(2 Chronicles 16:1 & 2 Chronicles 15:19) →

for some p^•, q^•, we have p^•(36) = q^•(35).

Suppose p^• = 10082.

 

For those wondering how we get p^• = 10082: We have some point p^• such that the 36^th year from p^• (in 2 Chronicles 16:1) and the 35^th year from q^• = p^• + 1 (in 2 Chronicles 15:19) fall somewhere during the reign of 3p_I≡Baash; i.e., between 10104 until 10127. 2 Chronicles 14:1 tells us there was quiet ten years after Asa began to reign. We've already estimat- ed that point p^• must be at least ten years prior to Asa; so now we're look- ing at an additional 10 years after he began to reign, i.e., at least 20 or more years before Asa. The second king of Judah, 2q≡Abijah, reigned for 3 years and the first king of Judah, 1q≡Rehoboam, ruled for 17 years. So this point p^• must rest on 10082, if not before; i.e., when the Israelites split up into the nation of Israel and that of Judah and 1q≡Rehoboam began to reign.

 

To harmonize 2 Chronicles 16:1, 1 Kings 15:28 and 1 Kings 15:33 we look at p^• =1p≡Rehoboam = 10082.

1q(35) = ((10082 + 35) = 10117

10117 puts us in the range of 1 Kings 15:33; i.e.,

3p_I≡(10104 until 10127)≡Baash and

we have harmony with 2 Chronicles 16:1

 

1q(35) = 10117 → for some n^•

3q(n^• ) = (10102 + n^•) = 10117 → n^• = 15.

Note that n^• = 15 ↔ 3p = 10102

 

Here we've shown that Asa≡3p(15) = 10117; i.e., the 15^th year in the reign of Asa is the 35^th year when counting from Rehoboam≡1p. In Asa's 15^th year there was war between Asa and Baasha (1 Kings 15:32).

 

We find corroboration in 2 Chronicles 15: ⁹And he (Asa) gathered all Judah and Benjamin, and the strangers with them out of Ephraim and Manasseh, and out of Simeon: for they fell to him out of Israel in abun- dance, when they saw that the LORD his God was with him. ¹⁰So they gathered themselves together at Jerusalem in the third month, in the fifteenth (= n^•) year of the reign of Asa.

And n^• = 15 is the fifteenth year of the reign of Asa.

n^• = 15 ↔ 3p = 10102 verifies our assumption from 1 Kings 15:10; i.e., that 3p = 10102 is true.

 

Recall that from 2 Chronicles 15:19 we got

1q(35) = 10117 → for some n^•,

3q(n^• ) = (10102 + n^•) = 10117 → n^• = 15.

Therefore in order to harmonize the fifteenth (= n^•) year of 2 Chronicles 15:10 with all the other verses regarding Asa, we must have n^• = 15 and only 3p = 10102 will give us n^• = 15. The only other possibility from 1 Kings 15:9 is 3p =10101; but then n^• = 16. Therefore n^• = 15 ↔ 3p = 10102, which we assumed in 1 Kings 15:10.

 

We now have:

3q≡Asa reigned for 41 years: from 10102 until 10143 (= 3Q).

2p_I≡Nadab reigned for 2 years: from 10103 until 10104 (= 2P_I)

3p_I≡Baash reigned for 24 years: from 10104 until 10127 (= 3P_I).

 

On November 8 we look at Elah, Zimri, Tibni, and Omri

ELAH, ZIMRI, TIBNI and OMRI

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 5:

3q≡Asa reigned for 41 years: from 10102 until 10143 (= 3Q).

3p_I≡Baash reigned for 24 years: from 10104 until 10127 (= 3 P_I).

 

(The green is for anyone needing help in understanding the formal argu- ments in aqua.)

 

1 Kings 16: ⁶So Baasha slept with his fathers, and was buried in Tirzah: and Elah his son reigned in his stead. →

Elah≡4p_I ≤ Baasha≡3P_I = 10127.

 

We're counting from P_I regarding the Israeli kings (see November 3).

 

1 Kings 16: ⁸In the twenty and sixth year of Asa king of Judah began Elah the son of Baasha to reign over Israel in Tirzah, two years →

4p_I ≤ 10127 = (3p(26) ∆ 3q(26)) →

4p_I ≤ 10127 = (10102 – 1) + 26) ∆ (10102 + 26) →

4p_I = 10127 = (10127 ∆ 10128)

4p_I = 3P_I = 10127 = 3p(26)

1 Kings 16:8 agrees with 1 Kings 16:6.

 

The first three lines in aqua show that Elah began to reign either in 10127 or 10128. But Baasha≡3P_I = 10127; and Elah≡4p_I wouldn't begin to reign a year after Baasha≡3P_I; i.e., Elah wouldn't begin to reign in 10128. Therefore 4p_I = 3P_I = 10127.

 

(4p_I(2) = 4P_I) →

4p_I(2) = 4P_I = ((10127 – 1) + 2) = 10128 = 4P_I.

 

We're counting from P_I regarding the Israeli kings (see November 3).

 

1 Kings 16: ¹⁰And Zimri went in and smote him, and killed him (Elah: 1 Kings 16:8), in the twenty and seventh year of Asa king of Judah, and reigned in his stead →

Zimri≡5p_I = 4P_I

 

Because Zimri killed Elah, we know Zimri began to rule that same year.

 

4P_I = 10128 = Asa≡(3p(27) ∆ 3q(27)) →

4P_I = 10128 = ((10102 – 1) + 27) ∆ (10102 + 27) →

4P_I = 10128 = (10128 ∆ 10129).

4P_I = 10128 = 3p(27) 1 Kings 16:10 agrees with 1 Kings 16:8.

4p_I≡Elah reigned for 2 years: from 10127 until 10128 (= 4P_I) and

5p_I = 10128.

 

The first four lines of aqua confirm that Elah's reign ended in 10128 (ac- cording to 1 Kings 16:8); and that Elah's reign ended in the 27^th year of Asa (according to 1 Kings 16:10), when counting from 3p; so 1 Kings 16:8 agrees with 1 Kings 16:10. 1 Kings 16:8 give Elah's reign as 2 years and because Zimri≡5p_I killed Elah (1 Kings 16:10), we know Zimri≡5p_I began to reign that same year; therefore 5p_I = 4P_I.

 

1 Kings 16: ¹⁵In the twenty and seventh year of Asa king of Judah did Zimri reign seven days in Tirzah. And the people were encamped against Gibbethon, which belonged to the Philistines. ¹⁶And the people that were encamped heard say, Zimri hath conspired, and hath also slain the king: wherefore all Israel made Omri, the captain of the host, king over Israel that day in the camp. & (1 Kings 16:10) →

3p(27) = 5p_I= 5P_I = Omri≡6p_I = 10128

 

In 1 Kings 16:10 and 1 Kings 16:15 Zimri's reign began in Asa's 27^th year when counting from 3p. Because Zimri reigned only seven days, we know that 5p_I= 5P_I. In 1 Kings 16:16 Israel made Omri king; therefore 5p_I = 6p_I. 1 Kings 16:10 gives the year as 5p_I = 10128.

5p_I≡Zimri reigned for 7 days: from 10128 until 10128 (= 5P_I).

 

1 Kings 16: ²¹Then were the people of Israel divided into two parts: half of the people followed Tibni the son of Ginath, to make him king; and half followed Omri. ²²But the people that followed Omri prevailed against the people that followed Tibni the son of Ginath: so Tibni died, and Omri reigned. ²³In the thirty and first year of Asa king of Judah began Omri to reign over Israel, twelve years: six years reigned he in Tirzah →

Omri≡6p_I(6) = Asa≡(3p(31) ∆ 3q(31)) →

6p_I(6) = (10102 – 1) + 31) ∆ (10102 + 31) →

Omri≡6p_I(6) = (10132 ∆ 10133) &

6p_I(6) = (10128 – 1) + 6 = 10133.

Omri≡6p_I(6) = 3q(31) when 6p_I = 10128

1 Kings 16:23 and 1 Kings 16:15-16 agree.

 

According to 1 Kings 16:23 Omri reigned for six years in Tirzah, then over all of Israel beginning in the 31^st year of Asa; therefore we'd expect Omri≡6p_I(6) = Asa≡(3p(31) ∆ 3q(31)). The third line of aqua shows that, when 6p_I = 10128, then the 6^th year of Omri, when counting from p_I, is the 31^st year of Asa, when counting from 3q. Therefore 1 Kings 16:15-16 and 1 Kings 16:23 agree.

 

Tibni≡7p_I = 10128 & (6p_I = 7p_I) →

6p_I(6) = 7p_I(6) →

Tibni≡7p_I(6) = 7P_I = 10133.

7p_I≡Tibni reigned for 6 years: from 10128 until 10133 (=7P_I).

 

In 1 Kings 16:21 Tibni and Omri began to reign the same year. Therefore Tibni≡7p_I = 6p_I and according to our corollary (p = p^• ↔ p(n) = p^•(n)), 6p_I(6) = 7p_I(6). So after six years in Tirzah Omri ruled over all Israel when Tibni died in 10133 = 6p_I(6) (1 Kings 16:22).

 

(1 Kings 16:23) → Omri≡6p_I(12) = ((10128 – 1) + 12) = 10139 = 6P_I

6p_I≡Omri reigned for 12 years: from 10128 until 10139 (= 6P_I)

 

On October 10 we look at Ahab, Jehoshaphat, and Ahaziah.

AHAB, JEHOSHAPHAT and AHAZIAH

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 8:

 

3q≡Asa reigned for 41 years: from 10102 until 10143 (= 3Q).

6p_I≡Omri reigned for 12 years: from 10128 until 10139 (= 6P_I)

 

(The green is for anyone needing help in following the formal arguments in aqua.)

 

1 Kings 15:²⁴And Asa slept with his fa- thers, and was buried with his fathers in the city of David his father: and Jehoshaphat his son reigned in his stead. →

4p≡Jehosphaphat ≤ 3Q = 10143.

 

1 Kings16:²⁸So Omri slept with his fathers, and was buried in Samaria: and Ahab his son reigned in his stead →

Ahab≡8p_I ≤ Omri≡6P_I = 10139

 

(7p_I was Tibni, who also ruled during the first six years of Omri's reign. See November 8)

 

1 Kings 16:²⁹And in the thirty and eighth year of Asa king of Judah began Ahab the son of Omri to reign over Israel: and Ahab the son of Omri reigned over Israel in Samaria twenty and two years →

8p_I ≤ 10139 = Asa≡3p(38) ∆ 3q(38) →

8p_I ≤ 10139 = (10102 – 1) + 38) ∆ (10102 + 38) →

8p_I = 10139 = (10139 ∆ 10140).

8p_I = 10139 = 3p(38).

1 Kings 16:29 agrees with 1 Kings 16:28.

 

We use the formulas p(n) = ((x – 1) + n and q(n) = x + n (see October 29) to show that Ahab≡8p_I began to rule in the 38^th year of Asa when count- ing from Asa≡3p. So 1 Kings 16:29 agrees with 1 Kings 16:28. Because 1 Kings 16:28 implies, 8p_I ≤ Omri≡6P_I = 10139, we know 8p_I wouldn't be- gin to rule a year after 10139, i.e., in 10140. Therefore Ahab≡8p_I = 10139 = Asa≡3p(38).

Ahab≡8p_I(22) = ((10139 – 1) + 22) = 10160.

8p_I≡Ahab reigned for 22 years: from 10139 until 10160 (= 8P_I).

 

We're using the formula p(n) = ((x – 1) + n) and counting from P_I regarding the Israeli kings (see November 3).

 

1 Kings 22:⁴¹And Jehoshaphat the son of Asa began to reign over Judah in the fourth year of Ahab king of Israel. ⁴²Jehoshaphat was thirty and five years old when he began to reign; and he reigned twenty and five years in Jerusalem… →

Jehoshaphat≡4p = Ahab≡(8p_I(4) ∆ 8q_I (4)) →

4p = ((10139 – 1) + 4) ∆ (10139 + 4)) →

4p = (10142 ∆ 10143).

(1 Kings 15:10 & November 5) → Asa≡3q(41) = 3Q = 10143 →

4p = (10142 ∆ 3q(41)).

q(n) – 1 = q(n – 1) →

Asa≡3q(40) = 10142. Therefore,

4p = 3q(40) ∆ 4p = 3q(41).

 

Since 4p = (10142 ∆ 10143) and Asa≡3q(41) = 10143 we can substitute 3q(41) for 10143 in 4p = (10142 ∆ 10143). 

So 4p = (10142 ∆ 3q(41)).

Since 3q(41) = 10143, we can subtract 1 from each side to get 3q(41) – 1 = 10143 – 1. Using the identity, p(n) ± m = p(n ± m) ((see October 29) and letting m = 1, we get q(n) – 1 = q(n – 1). So 3q(41 – 1) = 3q(40) = 10142. Replacing 10142 with 3q(40) in 4p = (10142 ∆ 3q(41)), we get 4p = 3q(40) ∆ 4p = 3q(41), where 3q(41) = 3Q.

 

2 Chronicles 16:¹² And Asa in the thirty and ninth year of his reign was diseased in his feet, until his disease was exceeding great: yet in his dis- ease he sought not to the LORD, but to the physicians. ¹³And Asa slept with his fathers, and died in the one and fortieth year of his reign. →

Conclude 4p = 3q(40).

 

Asa would have set up his son, 4p≡Jehoshaphat, to reign in Asa≡3q(40) rather than Asa≡3q(41) (= 3Q), the year of his death. First of all, the as- sumption of Divinity prohibits Scripture from including Asa's foot prob- lems incidentally. The feet and hands symbolize someone's will. That Asa was diseased in his feet, suggests he no longer had the will to rule or fol- low God's rule, at least not by himself. Secondly, that Asa sought not...the LORD, but...the physicians suggests he would have turned to Jehosha- phat (rather than to the LORD) before his last year. So in order to pro- ceed we define Jehoshaphat≡4p = 10142 (= 3q(40)) rather than 10143. If our assumption contradicts anything, we can always make the necessary changes.

(1 Kings 22:41-42) →

Jehoshaphat≡4Q = 4q(25) →

4Q = ((10142 + 25) →

4Q = (10167).

So 4q≡Jehoshaphat ruled for 25 years: from 10142 until 10167 (= 4Q).

 

1 Kings 22:⁵⁰And Jehoshaphat slept with his fathers, and was buried with his fathers in the city of David his father: and Jehoram his son. ⁵¹Ahaziah the son of Ahab began to reign over Israel in Samaria the seventeenth year of Jehoshaphat king of Judah, and reigned two years over Israel →

Ahaziah≡9p_I = Jehoshaphat≡(4p(17) ∆ 4q(17) →

9p_I = (10142 – 1) + 17) ∆ (10142 + 17) →

9p_I = (10158 ∆ 10159) →

9p_I(2) = 9 P_I = (10159 ∆ 10160)

 

Again we use the identity, p(n) ± m = p(n ± m), and corollary p = p(1) (see October 29). We let m = 1 and 9p_I = 9p_I(1); so if 9p_I = 10158 then 9p_I + 1 = 10158 + 1 and 9p_I(1+ 1) = 9p_I(2) = 10159. Similarly if 9p_I = 10159 then 9p_I(2) = 10160.

From 1 Kings 16:29 we know

8p_I≡Ahab reigned for 22 years: from 10139 until 10160 (= 8PI)

1 Kings 22:⁴⁰So Ahab slept with his fathers; and Ahaziah his son reigned in his stead. & 1 Kings 16:29 →

Ahaziah≡9p_I ≤ 10160 (= 8P_I)≡Ahab.

Therefore 9P_I = 10159 contradicts 1 Kings 22:40 and

9P_I = 10160 = 8P_I.

9p_I(2) = 10160 →

9p_I = 10159 = 4q(17).

9p_I≡Ahaziah reigned for 2 years: from 10159 until 10160 (= 9P_I).

 

Again we use the identity p(n) ± m = p(n ± m), m = 1 to go from 9p_I(2) = 10160 to 9p_I = 10159. From 1 Kings 22:50 we got 4q(17) = 10159 =9p_I.

 

On November 12 we look at Joram, Jehoram, and Ahaziah.

Joram, Jehoram, and Ahaziah

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 10:

9p_I≡Ahaziah reigned for 2 years: from 10159 until 10160 (= 9P_I).

4q≡Jehoshaphat ruled for 25 years: from 10142 until 10167 (= 4Q).

(1 Kings 22:51 & 1 Kings 22:40) → 9p_I = 4q(17).

 

(The green is for anyone needing help in following the formal arguments in aqua.)

 

2 Kings 1:¹⁷So he (Ahaziah, son of Ahab) died…And Jehoram (Joram) reigned in his stead in the second year of Jehoram the son of Jehoshaphat king of Judah; because he had no son.

Since 9p_I(2) = 9P_I, we let Joram≡10p_I = 10160.

Joram≡10p_I = Jehoram≡(5p(2) ∆ 5q(2))

 

We have here a Jehoram≡10p_I of Israel and a Jehoram≡5p of Judah, both reigning at the same time. They both ruled for 12 years with only 1 year difference between 10p_I and 5p. To avoid confusion we let Jehoram≡10p_I of Israel be Joram≡10p_I (they're one and the same name). Since 9p_I≡Aha- ziah reigned only two years we set Joram≡10p_I = 9P_I.

 

2 Kings 3:1: Now Jehoram (Joram) the son of Ahab began to reign over Israel in Samaria the eighteenth year of Jehoshaphat king of Judah, and reigned twelve years. →

Joram≡10p_I(12) = 10P_I = (10160 – 1) + 12) = 10171.

10p_I≡Joram reigned for 12 years: from 10160 until 10171 (=10P_I).

 

We're using the formula p(n) = ((x – 1) + n) and counting from P_I re- garding the Israeli kings (see October 29 and November 3).

 

Joram≡10p_I = 10160 = Jehoshaphat≡(4p(18) ∆ 4q(18))

From November 10 and 2 Chronicles 16:12-13 we defined

4p = 3q(40) = 10142 where 3q≡Asa. Therefore

10p_I = 10160 = (10142 – 1) + 18) ∆ (10142 + 18)

10p_I = 10160 = (10159 ∆ 10160)

10p_I = 10160 = 4q(18)

2 Kings 3:1 agrees with 2 Kings 1:17 in that 10p_I = 4q(18) =10160 and both verses harmonize with 2 Chronicles 16:12-13 in that 4p = 10142.

 

On November 10 we determined that Asa would have set up his son, 4p≡Jehoshaphat, to reign in Asa≡3q(40) rather than Asa≡3q(41) = 3Q, the year of his death. Therefore we defined Jehoshaphat≡4p = 10142. See October 29 for the formulas p(n) = (x – 1) + n and q(n) = x + n.

 

1. (2 Kings 1:17 & 2 Kings 3:1)→
2. 10p_I = 10160 &
3. (5p(2) ∆ 5q(2)) = 4q(18).
4. 5p(2) = 4q(18) → 5p = 4q(17)
5. 5q(2) = 4q(18) → 5q = 4q(17) → 5p = 4p(17)
6. 5p = (4p(17) ∆ 4q(17))
7. ((1 Kings 22:50-51 & 1 Kings 22:40) & 2 Kings 3:1) →
8. we're counting from 4q to define 9p_I & 10p_I respectively →
9. 5p = 4q(17) = 10159.

 

Line 1: 2 Kings 1:17 implies Joram≡10p_I = Jehoram≡(5p(2) ∆ 5q(2)).

2 Kings 3:1 gave us Joram≡10p_I = 4q(18).

In line 3 we can let 5p(2) ∆ 5q(2) = 4q(18).

In line 4 we suppose that 5p(2) = 4q(18). If we subtract 1 from each side, 5p(2) – 1 = 4q(18) – 1, we get 5p = 4q(17) (see October 29 where we define the identity: p(n) ± m = p(n ± m).

In line 5 we suppose 5q(2) = 4q(18). Again we subtract 1 from each side and get 5q = 4q(17), which implies 5p = 4p(17) (see October 29 and our corollary: p(n) = p^•(m) ↔ q(n) = q^•(m)).

We can now conclude in line 6 that 5p = (4p(17) ∆ 4q(17).

On November 10 we looked at 1 Kings 22:40 and 1 Kings 22:⁵¹Ahab be- gan to reign over Israel in Samaria the seventeenth year of Jehoshaphat king of Judah. We then determined that 9p_I = 10159 = 4q(17). So 1 Kings 22:51 refers to 4q. Also 2 Kings 3:1 refers to 4q such that 10p_I = 4q(18). Since in line 2 we began with 10p_I, we conclude in line 7 and 8 that we're counting from 4q.

Therefore in line 9 we have 5p = 4q(17) = 10159 (since we already know 10160 = 4q(18)).

 

2 Kings 8:¹⁶And in the fifth year of Joram the son of Ahab king of Israel, Jehoshaphat being then king of Judah, Jehoram the son of Jehoshaphat king of Judah began to reign. ¹⁷Thirty and two years old was he when he began to reign; and he reigned eight years in Jerusalem. →

Jehoram≡5p = Joram≡(10p_I(5) ∆ 10q_I(5)) →

5p = ((10160 – 1) + 5) ∆ (10160 + 5) →

5p = (10164 ∆ 10165): since this contradicts

2 Kings 3:1 → 5p = 10159, we have,

2 Kings 8:16-17→ that for some n^•,

(5p(n^•) ∆ 5q(n^•)) = (10164 ∆ 10165).

 

The first two lines in aqua is business as usual. But then in lines 3 and 4 we run into a contradiction with 2 Kings 3:1. 5p couldn't begin to reign in 10159 and 10164 or 10165. (Recall that the word, "and" or & means both must be true; see November 1). Therefore in line 5 we conclude that for some reason 2 Kings 8:16-17 refers to a later date; i.e., n^• years after 10159, where 5 ≤ n^• ≤ 7.

 

1 Kings 22:51 & 1 Kings 22:40 → we count from 4q(17) to define 9p_I.

2 Kings 3:1→ we count from 4q(18) to define 10p_I,

So in 2 Kings 8:16-17 we count from 5q(n^•) to define 10p_I(5).

Therefore 5q(n^•) = 10p_I(5) = (10164 ∆ 10165).

 

Since we already have (5p(n^•) ∆ 5q(n^•)) = (10164 ∆ 10165), and we just determined that we count from 5q to determine 10p_I(5), we have

5q(n^•) = 10p_I(5) = (10164 ∆ 10165).

 

2 Kings 3:1 → 10p_I = 10160 & 2 Kings 8:16 →

10p_I(5) = ((10160 – 1) + 5) = 10164 →

5q(n^•) = 10164.

2 Kings 3:1 → 5p = 10159 → 5q = 10160 & n^• = 5.

 

2 Kings 8:16 refers to the fifth year of 10p_I, and 10p_I(5) = 10164. Since we're counting from 5q, we get 5q(n^•) = 10164. Since 5p = 10159, we have 5q = 10160. Adding 4 to both sides of the latter equality we get 10164 = 5q + 4 = 5q(1 + 4) = 5q(5). (Recall that q + n = q(1 + n); see October 29) So n^• = 5.

 

2 Chronicles 21:¹Now Jehoshaphat slept with his fathers, and was buried with his fathers in the city of David. And Jehoram his son reigned in his stead.

 

2 Chronicles 21:⁴Now when Jehoram (son of Jehoshaphat) was risen up to the kingdom of his father, he strengthened himself, and slew all his brethren with the sword, and divers also of the princes of Israel.…⁵Jeho- ram was thirty and two years old when he began to reign, and he reigned eight years in Jerusalem →

 

2 Kings 8:16-17: → 5q(n^•)) = 10164 is when 5q(5) (son of Jehoshaphat) was risen up to the kingdom of his father.

 

Now we know why 2 Kings 8:16-17 gave us 10164 instead of 10159. In 10159 Jehoram shared the throne with his father. In 10164 he knocked off all his brethren to show who's boss.

 

2 Chronicles 21:²⁰Thirty and two years old was he (Jehoram) when he began to reign, and he reigned in Jerusalem eight years, and departed without being desired. Howbeit they buried him in the city of David, but not in the sepulchres of the kings. →

(2 Chronicles 16:12-13) → Jehoshaphat≡4Q = (10167).

(2 Kings 3:1) → Jehoram≡5p = 10159 →

5p reigned as co-regent with his father, Jehoshaphat, for 9 years (10159 to 10167).

(2 Kings 8:16-17) →

5q(5) = 10164 is when some kind of bloody change occurred.

(2 Kings 8:16-17 & 2 Chronicles 21:4, 5 & 2 Chronicles 21:20) →

Jehoram ruled for 8 years beginning 5q(5) = 10164 →

Jehoram≡5Q = 10171 ( = 10164 – 1 + 8)

2 Kings 3:1 → Jehoram reigned twelve years, and (10171 – 10159) = 12 so we have agreement with 2 Kings 3:1.

5q≡Jehoram reigned for 12 years: from 10159 until 10171 (= 5Q).

2 Kings 1:16-17→ 10p_I = (5p(2) ∆ 5q(2)) = 10160

Since 5p(2) = (10159 – 1) + 2 = 10160, the results of 2 Kings 1:16-17 agrees with 2 Chronicles 21:20. However, while 1 Kings 22:51 & 1 Kings 22:40 defines 9p_I from 4q, and 2 Kings 3:1 defines 10p_I from 4q, 2 Kings 1:17 defines 10p_I from 5p(2)—not 5q(2).

 

2 Chronicles 22:¹And the inhabitants of Jerusalem made Ahaziah his youngest son king in his stead: for the band of men that came with the Arabians to the camp had slain all the eldest. So Ahaziah the son of Je- horam king of Judah reigned. →

Ahaziah≡6p = 5Q≡Jehoram= 10171.

 

Since 5q knocked off all the familial competition in 2 Chronicles 21:4, we doubt 6p is less than 5Q.

 

2 Kings 8:²⁵In the twelfth year of Joram the son of Ahab king of Israel did Ahaziah the son of Jehoram king of Judah begin to reign. ²⁶Two and twenty years old was Ahaziah when he began to reign; and he reigned one year in Jerusalem. →

10p_I(12) = 6p≡Ahaziah

2 Kings 9:²⁹…in the eleventh year of Joram the son of Ahab began Aha- ziah to reign over Judah →

10q_I(11) = 6p≡Ahaziah

 

Note that we mustn't confuse 9p_I≡Ahaziah with 6q≡Ahaziah of 2 Kings 8:25 and 2 Kings 9:29.

 

(2 Kings 3:1) → 10p_I(12) = 10171.

(10p_I(12) = 10q_I(11) = 10171 so 2 Kings 3:1 and 2 Kings 8:29 agree.

2 Kings 8:25, 26→

Ahaziah≡6q = 10172 = 6Q.

2 Kings 8:25-26 & 2 Kings 9:29 agree with 2 Chronicles 22:1

6q≡Ahaziah reined for 1 year: from 10171 until 10172 (= 6Q).

 

Recall that on November 1 we looked at (2 Kings 8:25 & 2 Kings 9:29) → 6p = 10p_I (12) = 10q_I (11) and planned to check it out once we deter- mined the numbers. 2 Kings 3:1 gives us 10p_I(12) = 10171 and 2 Chron- icles 22:1 gives us, 6p = 10171; so we do indeed have agreement.

 

On November 15 we look at Joram (same guy again), Jehu and Athaliah

Joram, Jehu and Athaliah

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 12:

6q≡Ahaziah reined for 1 year: from 10171 until 10172 (= 6Q).

10p_I≡Joram reigned for 12 years: from 10160 until 10171 (=10P_I).

 

2 Kings 8:²⁶Two and twenty years old was Ahaziah (≡6q) when he began to reign; and he reigned one year in Jerusalem. And his mother's name was Athaliah, the daughter of Omri king of Israel.

2 Chronicles 22: ²Forty and two years old was Ahaziah (≡6q) when he began to reign, and he reigned one year in Jerusalem. His mother's name also was Athaliah the daughter of Omri.

 

In these two verses we have here the same guy, Ahaziah (≡6q), but a twenty year difference between the ages when he supposedly began to reign in 10171 (see November 12). If we subtract 22 from 10171 we get 10149. If we subtract 42 from 10171 we get 10129 = Omri≡6q_I (10128 to 10139; see November 8). While 2 Kings 8:26 refers to the chronological age of Ahaziah≡6q, the 42 years of 2 Chronicles 22:2 refers to the dynas- ty of Omri≡6q_I.

 

2 Kings 8:26 and 2 Chronicles 22:2 have the same grammatical construc- tion as Samuel 13 (YLT) ¹A son of a year is Saul in his reigning, yea, two years he hath reigned over Israel (see October 29). Saul's dynasty began with himself; therefore Saul reigned one year from the dynasty of Saul and two years from point p. After him and David the kings of Judah fol- lowed David's lineage—except in the case of Athaliah as described be- low. Ahaziah reigned 42 years from the dynasty of Omri≡6q_I. 2 Kings 8:26 and 2 Chronicles 22:2 could have been translated as, A son of 22 (42) years was Ahaziah when he began to reign.

 

But now we wonder why 2 Chronicles 22:2 relates Ahaziah≡6q, a king of Judah, to Omri≡6q_I, a previous king of Israel.

 

Omri≡6q_I had a son, Ahab≡8p_I (10139 to 10160; see November 10). Ahab≡8p_I had a son, Joram≡10p_I (10160 to 10171 see November 12) and a daughter, Athaliah≡7q. Because Ahab≡8p_I was a son of Omri≡6p_I, 2 Kings 8:26 records Athaliah≡7q as a daughter of Omri≡6p_I (10128 to 10139). Being a son or daughter of x doesn't necessary mean an immedi- ate offspring. Athaliah≡7q and Jehoram≡5p (10159 to 10171; see No- vember 12) had a son, Ahaziah≡6q (see November 12): ²⁵Ahaziah the son of Jehoram king of Judah (2 Kings 8). Thus the house of Ahab≡8p_I in- cluded Ahaziah≡6q, who ruled in Judah from 10171 until 10172, which is where we left off on November 12.

 

In 2 Kings 8: ²⁷…he (Ahaziah≡6q) walked in the way of the house of Ahab, and did evil in the sight of the LORD, as did the house of Ahab: for he was the son in law (i.e., related by marriage) of the house of Ahab.

 

Thus while Ahaziah≡6q was in the bloodline of David (through the father, Jehoram≡5p), 2 Chronicles 22:2 (and 2 Kings 8:27) also relates Ahazi- ah≡6q to the house of Ahab≡8p_I (through the mother, Athaliah≡7q).

 

While Joram≡10p_I (10160 to 10171 see November 12), son of Ahab≡8p_I, still ruled in Israel, God had Jehu≡11q_I anointed king over Israel in 2 Kings 9. God used Jehu≡11q_I to wipe out the whole house of Ahab≡8p_I (son of Omri≡6p_I). ⁸For the whole house of Ahab shall perish: (2 Kings 9). Since the house of Ahab included Ahaziah≡6q, God had Jehu≡11q_I eliminate him and his family as well:

⁷…the destruction of Ahaziah (≡6q) was of God by coming to Joram: for when he was come, he went out with Jehoram (Joram≡10p_I) against Jehu the son of Nimshi, whom the LORD had anointed to cut off the house of Ahab. ⁸And it came to pass, that, when Jehu was executing judgment up- on the house of Ahab, and found the princes of Judah, and the sons of the brethren of Ahaziah, that ministered to Ahaziah, he slew them (2 Chron- icles 22:7-8).

 

We read of the death of Joram≡10p_I and Ahaziah≡6q in 2 Kings 9:²¹And Joram (≡10p_I) said, Make ready. And his chariot was made ready. And Joram king of Israel and Ahaziah (≡6q) king of Judah went out, each in his chariot, and they went out against Jehu, and met him in the portion of Naboth the Jezreelite. ²⁴And Jehu drew a bow with his full strength, and smote Jehoram between his arms, and the arrow went out at his heart, and he sunk down in his chariot. ²⁷…when Ahaziah the king of Judah saw this, he fled by the way of the garden house. And Jehu followed after him, and said, Smite him also in the chariot. And they did so at the going up to Gur, which is by Ibleam. And he fled to Megiddo, and died there →

 

From November 12:

(2 Kings 8:29) →Ahaziah≡6q(1) = 6Q = 10172 &

(2 Kings 3:1) → Joram≡10p_I(12) = 10P_I = (10160 – 1) + 12) = 10171

(2 Kings 9:21: & 2 Kings 9:24 & 2 Kings 9:27) →

Ahaziah≡6Q = 10P_I≡Joram.

But Joram≡10P_I = 10171 ≠ Ahaziah≡6Q = 10172.

So we have an apparent contradiction and must conclude.

~(6Q = 10P_I).

 

By using our identity, p(n) ± m = p(n ± m), and corollary, p(n + 1) = q(n), from October 29:

10p_I(12) + 1 = 10p_I(12 + 1) = 10q_I(12) = 10171 + 1 = 10172 = 10Q_I.

Joram≡10Q_I = 10172 = Ahaziah≡6Q.

So we're suddenly counting from 10q_I in respect to a king of Israel.

We go back to 2 Kings 3:1 from November 12 and make the correction:

10p_I≡Joram reigned for 12 years: from 10160 until 10171 (=10P_I)

10q_I≡Joram reigned for 12 years: from 10160 until 10172 (= 10Q_I)

 

2 Kings 9:¹³Then they hasted, and took every man his garment, and put it under him on the top of the stairs, and blew with trumpets, saying, Jehu is king….→

Jehu≡11p_I = 10172

2 Kings 10:³⁵…Jehu slept with his fathers: and they buried him in Samar- ia. And Jehoahaz his son reigned in his stead. ³⁶And the time that Jehu reigned over Israel in Samaria was twenty and eight years &

Jehu≡11p_I = 10172 →

(11p_I(28) ∆ 11q_I(28)) →

((10172 – 1) + 28) ∆ (10172 + 28))

(10199 ∆ 10200).

 

Since we're now counting from q_I in respect to the Israeli kings we have,

11q_I≡Jehu reigned for 28 years: from 10172 until 10200 (= 11Q_I).

Jehoahaz≡12p_I ≤ 10200.

 

Meanwhile in Jerusalem, ¹…when Athaliah (≡7q) the mother of Ahaziah (≡6q) saw that her son was dead, she arose and destroyed all the seed royal. By whacking all contenders for the throne, Athaliah did God's will in wiping out the remainder of Ahaziah's line. ²But Jehosheba, the daugh- ter of king Joram (Jehoram≡5p), sister of Ahaziah, took Joash (Jehoash) the son of Ahaziah, and stole him from among the king's sons which were slain; and they hid him, even him and his nurse, in the bedchamber from Athaliah, so that he was not slain.³And he was with her hid in the house of the LORD six years. And Athaliah did reign over the land (2 Kings 11). Not in David's lineage, Athaliah had no right to the throne.

 

We wonder why God allowed Joash to escape after he ordered the whole house of Ahab destroyed. In 2 Kings 8:¹⁹…the LORD would not destroy Judah for David his servant's sake, as he promised him to give him alway a light, and to his children. Therefore God spared Joash so that the royal line of David could continue to Christ.

 

Six years later, in 2 Kings 11, Jehoiada the priest ¹²…brought forth the king's son (Joash, or Jehoash) and put the crown upon him…and they made him king… ²⁰…all the people of the land rejoiced…and they slew Athaliah with the sword beside the king's house →

Athaliah≡7p = 10172

Athaliah≡(7q(6)) = 10172 + 6 = 10178 = 7Q.

7q≡Athaliah reigned for 6 years; from 10172 until 10178 (= 7Q).

8p≡Joash (Jehoash) = 10178

 

On November 17 we look at Jehoash, Jehoahaz, Amaziah and Joash.

Jehoash, Jehoahaz, Amaziah and Joash

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 15:

2 Kings 10:36 →

11q_I≡Jehu reigned for 28 years: from 10172 until 10200 (= 11Q_I)

& 2 Kings 11:12, 20 → 8p≡Jehoash = 10178

 

2 Kings 12:¹In the seventh year of Jehu Jehoash began to reign; and forty years reigned he in Jerusalem…

2 Kings 11:12, 20 → 8p≡Jehoash = 10178

2 Kings 12:1 → Jehoash≡8p = 10178 = Jehu≡(11p_I(7) ∆ 11q_I(7))

2 Kings 10:36 → 11p_I≡Jehu = 10172. Therefore

8p = 10178 = ((10172 – 1) + 7) ∆ (10172 + 7)) →

8p = 10178 = (10178 ∆ 10179).

8p = 10178 = 11p_I(7)

2 Kings 11:12, 20 & Kings 12:1 & 2 Kings 10:36 agree.

2 Kings 12:1→ 8q(40) = 8Q &

2 Kings 11:12, 20 → 8p = 10178. Therefore

Jehoash≡8q(40) = (10178 + 40) = 10218.

8q≡Jehoash reigned for 40 years: from 10178 until 10218 (= 8Q)

 

2 Kings 13:¹In the three and twentieth year of Joash (8p≡Jehoash) the son of Ahaziah king of Judah Jehoahaz the son of Jehu began to reign over Israel…and reigned seventeen years.

2 Kings 12:1→ 8p≡Jehoash = 10178 &

Jehoash≡(8p(23) ∆ 8q(23) =

(10178 – 1) + 23) ∆ (10178 + 23) =

(10200 ∆ 10201), therefore

8p(23) = 10200

2 Kings 13:1 → Jehoahaz≡12p_I = Jehoash≡(8p(23) ∆ 8q(23), therefore

(2 Kings 12:1 & 2 Kings 13:1) → 12p_I = (10200 ∆ 10201)

2 Kings 10:36 → 11Q_I = 10200 →

12p_I ≤ 10200 = 8p(23), therefore

Because 12p_I would not begin to reign a year after 11Q_I, we conclude that 12p_I = 10200.

2 Kings 10:36 agrees with (2 Kings 12:1 & 2 Kings 13:1) →

12p_I = (10200 ∆ 10201).

(12p_I = 10200 & 2 Kings 13:1 → 12q_I(17) = 12Q_I) →

12q_I(17) = (10200 + 17) = 10217, therefore.

13p_I ≤ 12Q_I = 10217

12q_I≡Jehoahaz reigned for 17 years: from 10200 until 10217 (= 12Q_I).

 

2 Kings 13:⁹And Jehoahaz slept with his fathers; and they buried him in Samaria: and Joash (13q_I≡Joash) his son reigned in his stead. ¹⁰In the thirty and seventh year of Joash (8q≡Jehoash) king of Judah began Jeho- ash (13q_I≡Joash) the son of Jehoahaz to reign over Israel in Samaria, and reigned sixteen years.

 

Scripture interchanges the names Joash and Jehoash for the same guy. In 2 Kings 13:9 we have a king of Israel, Joash (13q_I≡Joash). In 2 Kings 13:10 he becomes Jehoash (13q_I≡Joash), while at the same time 2 Kings 13:10 refers to a different Joash (8q≡Jehoash; see 2 Kings 12:1) ruling in Jerusalem. So as not to confuse 8q≡Jehoash (or Joash) in Judah with Jo- ash≡13q_I (or Jehoash) in Israel, we always write Jehoash (e.g., ≡8q) for a king in Jerusalem and Joash (e.g., ≡13q_I) for one in Samaria.

 

2 Kings 13:1 → 12Q_I = 10217

Kings 13:9 → Joash≡13p_I ≤ 12Q_I = 10217

2 Kings 13:10 →13p_I = Jehoash≡(8p(37) ∆ 8q(37))

2 Kings 12:1→ Jehoash≡8p = 10178. Therefore

Joash≡13p_I = ((10178 – 1) + 37) ∆ (10178 + 37) →

2 Kings 13:10 →13p_I = (10214 ∆ 10215)

In the case of 12p_I we had ((2 Kings 12:1 & 2 Kings 13:1) →

12p_I = (10200 ∆ 10201)) & 8p(23) = 10200 &

2 Kings 10:36 → 11Q_I = 10200 → 12p_I ≤ 10200 = 8p(23), which made it easy to choose 12p_I = 8p(23)

Similarly in order to define 8p, 2 Kings 12:1 & 2 Kings 10:36 directed us to point 11p_I to get 8p = (11p_I(7)).

However with 13p_I we have no data directing us to point p in regards to a king of Judah. Instead we have (2 Kings 13:1) → 13p_I ≤ 12Q_I = 10217 &

2 Kings 13:10 →13p_I = (10214 ∆ 10215) where 8q(37) = 10215.

Because we're counting from a king of Judah without data directing us to 8p(37)—as in the other two cases—we let 13p_I = 8q(37).

Therefore 13p_I = 10215.

2 Kings 13:10 → (13p_I = 10215 & 13q_I(16) = 13Q_I), therefore

13q_I(16) = (10215 + 16)

13q_I(16) = 13Q_I = 10231

2 Kings 13:10 →

13q_I≡Joash reigned for 16 years: from 10215 until 10231 (=13Q_I)

2 Kings 13:1→ Joash≡13q_I ≤ 10217 = 12Q_I≡Jehoahaz →

for some n^•, Joash≡13q_I(n^•) = 10217 & n^• = 2.

 

2 Kings 14:¹In the second year of Joash (≡13q_I) son of Jehoahaz king of Israel reigned Amaziah the son of Joash (8q≡Jehoash) king of Judah. ²He was twenty and five years old when he began to reign, and reigned twenty and nine years in Jerusalem…→

2 Kings 13:9-10 → 13p_I = 10215

2 Kings 14:1 → Amaziah≡9p = (13p_I(2) ∆ 13q_I(2)), therefore

(2 Kings 13:9-10 & 2 Kings 14:1-2) →

9p = ((10215 – 1) + 2) ∆ (10215 + 2).

(2 Kings 13:9-10 & 2 Kings 14:1-2) → 9p = (10216 ∆ 10217).

2 Kings 14:2 → 9q(29) = 9Q

The data doesn't direct us to 9p = 13p_I(2). Though this time a king of Israel defines a king of Judah, we suspect that 9p = 13q_I(2) = 10217

2 Kings 12:1 → 8Q = 10218 therefore

9p ≤ 10218

2 Kings 12:1 agrees with (2 Kings 13:9-10 & 2 Kings 14:1-2).

 

2 Kings 14:¹⁶And Jehoash (Joash≡13q_I) slept with his fathers, and was buried… and Jeroboam his son reigned in his stead. ¹⁷And Amaziah the son of Joash (Jehoash≡8q) king of Judah lived after the death of Jehoash (13q_I≡Joash) son of Jehoahaz king of Israel fifteen years

(2 Kings 13:10 & 2 Kings 14:16) →

Jeroboam≡14p_I ≤ Joash≡13q_I(16) = 13Q_I

2 Kings 13:10 → 13Q_I = 10231, therefore

(2 Kings 14:16 & 2 Kings 13:10) →14p_I ≤ 10231.

2 Kings 14:1-2 → 9q(29) = 9Q

1 Kings 13:10 → 13q_I(16) = 13Q_I

2 Kings 14:16-17 → 9Q = 13Q_I + 15

Therefore 9q(29) = 13q_I(16) + 15

2 Kings 13:10 → 13q_I(16) = 10231

Therefore 9q(29) = 10231 + 15 = 10246 →

9q = 10246 – 28 = 10218 & 9p = 10217 & 9Q = 10246.

 

We were correct in assuming that 2 Kings 14:1 →

9p = 13q_I(2) = 10217. Similarly we should be correct in assuming 2 Kings 13:10 →13p_I = (10214 ∆ 10215) & 13p_I = 8q(37) = 10215.

 

9p≡Amaziah reigned for 29 years: from 10217 until 10246 (= 9Q).

 

(2 Kings 13:9-10 & 2 Kings 14:1-2) → 9p = (10216 ∆ 10217).

Since 2 Kings 14:16-17 → 9p = 10217,

(2 Kings 13:9-10 & 2 Kings 14:1-2) & 2 Kings 14:16-17 agree.

 

On November 19 we look at Jeroboam.

Jeroboam

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 17:

2 Kings 14:16 →

9q≡Amaziah reigned for 29 years: from 10217 until 10246(= 9Q).

2 Kings 13:10 →

13q_I≡Joash reigned for 16 years: from 10215 until 10231 (=13Q_I)

 

In 2 Kings 14 Joash≡13q_I, king of Israel, fought against Judah and took Amaziah≡9p captive: ¹³…Jehoash (13q_I≡Joash; see November 17) king of Israel took Amaziah king of Judah, the son of Jehoash the son of Aha- ziah, at Bethshemesh, and came to Jerusalem, and brake down the wall of Jerusalem from the gate of Ephraim unto the corner gate, four hundred cubits. ¹⁴And he took all the gold and silver, and all the vessels that were found in the house of the LORD, and in the treasures of the king's house, and hostages (including Amaziah), and returned to Samaria. Consequently ²¹…all the people of Judah took Azariah(≡10p), which was sixteen years old, and made him king instead of his father Amaziah.

 

Since Joash≡13Q_I = 10231, 2 Kings 14:13-21 occurred on or before 10231 < 10246 = 9Q≡Amiziah by at least 15 years. So at the age of 16, Azariah≡10p replaced Amiziah≡9p, now held captive in Samaria. So Az- ariah≡10p is less than 10246 by at least 15 years, probably more. So we define the letter a such that for some a, Azariah≡10q(a) = 10247.

 

The letter, a (as in alone), in Azariah≡10q(a), means 10q(a) = the year after they killed Amiziah and Azariah ruled by himself; i.e., a = q(a) – p is the number of years a king, p, reigned as co-regent with the previous king. Since 2 Kings 14:16 → 10246 = 9Q, we have 10q(a) = 10247.

 

2 Kings 14:²³In the fifteenth year of Amaziah the son of Joash (8q≡Jeho- ash; see November 17) king of Judah Jeroboam the son of Joash (≡13q_I) king of Israel began to reign in Samaria, and reigned forty and one years.

(2 Kings 14:16-17 → Amaziah≡9p = 10217

2 Kings 14:23 → Jeroboam≡14p_I(n^•) = (9p(15) ∆ 9q(15))); therefore

14p_I(n^•) = ((10217 – 1) + 15) ∆ (10217 + 15) = (10231 ∆ 10232).

2 Kings 13:10 → 10231 = 13Q_I.

Since (2 Kings 14:16-17 & 2 Kings 14:23) →

for some n^•, 14p_I(n^•) = 9q(15) = 10232,

n^• = a & 14q_I(a) = 10232.

(2 Kings 14:16-17 & 2 Kings 14:23) & 2 Kings 13:10 harmonize.

2 Kings 14:23 → 14q_I(41) = 14Q_I.

When we get to 2 Kings 15:1 we show that 14q_I < 13Q_I by at least 11 years.

 

2 Kings 14:²⁹And Jeroboam slept with his fathers, even with the kings of Israel; and Zachariah his son reigned in his stead →

Zachariah≡15p_I ≤ 14q_I(41) = 14Q_I

 

2 Kings 15:¹In the twenty and seventh year of Jeroboam king of Israel began Azariah son of Amaziah king of Judah to reign. ²Sixteen years old was he when he began to reign, and he reigned two and fifty years in Je- rusalem.

2 Kings 14:21→ 10q(a) = 10247.

If (2 Kings 14:21 & 2 Kings 15:1) → 14p_I(27) = 10q(a) = 10247, then 14p_I = 10247 – 26 = 10221.

If 14p_I = 13Q_I = 10231 (2 Kings 3:10) then

14q_I(27) = 10258 (according to 2 Kings 15:1).

If (2 Kings 14:21 & 2 Kings 15:1) → 14p_I(27) = 10q(a) = 10247, then 14p_I reigned as a co-regent with 13q_I for

a = 11 (= 10258 – 10247) years back from 13Q_I.= 10231.

So if 14q_I(11) = 10232 then 14q_I = 10232 – 10 = 10222 and again we have 14p_I = 10221.

2 Kings 14:23 → 14q_I(a) = 10232 verifies that (2 Kings 14:21 & 2 Kings 15:1) → 14p_I(27) = 10q(a) = 10247 is correct.

2 Kings 14:23 → 14q_I(41) = 14Q_I → 14Q_I = 10262.

14q_I≡Jeroboam reigned for 41 years: from 10221 until 10262 (= 14Q_I)

 

On November 22 we look at Azariah, Zachariah, Shallum, Menahem, Pekahiah, Pekah, & Jotham

Azariah, Zachariah, Shallum, Menahem, Pekahiah, Pekah, & Jotham

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 19:

2 Kings 15:1 → Jeroboam≡14Q_I = 10262.

2 Kings 15:2 → Azariah≡10q(52) = 10Q

2 Kings 15:⁷So Azariah slept with his fathers; and they buried him with his fathers in the city of David: and Jotham his son reigned in his stead.⁸In the thirty and eighth year of Azariah king of Judah did Zachariah the son of Jeroboam reign over Israel in Samaria six months. 

2 Kings 15:7 → Jotham≡11p ≤ 10Q

2 Kings 15:8 → Zachariah≡15q_I(.5) = 15Q_I; therefore

15p_I = 14Q_I (since 15p_I only reigned for six months.)

2 Kings 15:1 → 14Q_I = 10262

2 Kings 15:8 → 15p_I = (10p(38) ∆ 10q(38))

Because the data does not direct us to 10p(38), we assume

15p_I = (10q(38) (see November 17 under 2 Kings 13:9).

Therefore 10q(38) = 10262 &

10q = 10262 – 37 = 10225.

2 Kings 15:2 → 10q(52) = 10Q → 10Q = 10276

10q≡Azariah reigned for 52 years: from 10224 until 10276 (≡10Q)

2 Kings 15:¹³Shallum the son of Jabesh began to reign in the nine and thirtieth year of Uzziah (Azariah≡10p) king of Judah; and he reigned a full month in Samaria.

2 Kings 15:7 → 10q(38) = 10262 → 10q(39) = 10263

2 Kings 15:13 → Shallum≡16p_I = 10q(39) = 10263 &

16q_I(.0833) = 16Q_I (1/12 = .0833)

2 Kings 15:¹⁷In the nine and thirtieth year of Azariah king of Judah began Menahem the son of Gadi to reign over Israel, and reigned ten years in Samaria 

2 Kings 15:8 → 15p_I = (10q(38) &

(2 Kings 15:13 & 2 Kings 15:17) → 

Menahem≡17p_I = Shallum≡16p_I = (Zachariah≡15p_I + 1) = Azariah≡10q(39)) = 10263

15p_I≡Zachariah reigned for 6 months: from 10262 until 10262 (≡15Q_I)

16p_I≡Shallum reigned for 1 month: from 10263 until 10263(≡16Q_I)

Menahem≡17q_I(10) = (10263 + 10) = 17Q_I

2 Kings 15:17 →

17q_I≡Menahem reigned for ten years: from 10263 until 10273(≡17Q_I)

2 Kings 15:²²And Menahem slept with his fathers; and Pekahiah his son reigned in his stead. ²³In the fiftieth year of Azariah king of Judah Peka- hiah the son of Menahem began to reign over Israel in Samaria, and reigned two years.

2 Kings 15:22-23 → Pekahiah≡18p_I = (10p(50) ∆ 10q(50))

2 Kings 15:7-8 → Azariah≡10p = 10224; therefore

18p_I = ((10224 – 1) + 50) ∆ (10224 + 50)

(10p(50) ∆ 10q(50)) = (10273 ∆ 10274) →

2 Kings 15:17 & 2 Kings 15:22 → 10273 = 17Q_I & therefore

18p_I = 10273 (In this case the data directs us to point p; i.e., 10p)

2 Kings 15:23 → 18q_I(2) = 18Q_I = 10275.

18q_I≡Pekahiah reigned for 2 years: from 10273 until 10275 (≡18Q_I)

2 Kings 15:²⁵But Pekah the son of Remaliah, a captain of his, conspired against him, and smote him (Pekahiah) in Samaria…²⁷In the two and fiftieth year of Azariah king of Judah Pekah the son of Remaliah began to reign over Israel in Samaria, and reigned twenty years.

2 Kings 15:25 → Pekah≡19p_I = 18Q_I = 10275 (because Pekah slew Pe- kahiah, he would have begun to reign right away.)

Kings 15:7-8 → 10p = 10224.

2 Kings 15:27 → 19p_I = (10p(52) ∆ 10q(52)) &

(10p(52) = (10224 – 1) + 52 = 10275; therefore

(2 Kings 15:25 & 2 Kings 15:27) → 19p_I = 10275

(Kings 5:7-8 & 2 Kings 15:27) agree with 2 Kings 15:25.

2 Kings 15:27 → 19q_I(20) = 19Q_I = 10295

19q_I≡Pekah reigned for 20 years: from 10275 until 10295 (=19Q_I)

2 Kings 15:³⁰And Hoshea the son of Elah made a conspiracy against Pekah the son of Remaliah, and smote him, and slew him, and reigned in his stead, in the twentieth year of Jotham the son of Uzziah.

2 Kings 15:30 → Hoshea≡20p_I = 19Q_I = 10295

2 Kings 15:27 & 2 Kings 15:30 →

20p_I = (Jotham≡11p(20) ∆ Jotham≡11q(20)) = 10295 →

(11p ∆ 11q) = 10276

2 Kings 15:³² In the second year of Pekah the son of Remaliah king of Israel began Jotham the son of Uzziah king of Judah to reign. ³³Five and twenty years old was he when he began to reign, and he reigned sixteen years in Jerusalem. And his mother's name was Jerusha, the daughter of Zadok

2 Kings 15:32→ (11p ∆ 11q) = (19p_I(2) ∆ 19q_I(2)

2 Kings 15:25 & 2 Kings 15:27: → 19p_I = 10275 →

(11p ∆ 11q) = ((10275 – 1) + 2) ∆ (10275 + 2))

(11p ∆ 11q) = (10276 ∆ 10277)

2 Kings 15:7 → 10Q = 10276 → 11p ≤ 10276 →

(11p ∆ 11q) = 10276

2 Kings 15:30 → (11p ∆ 11q) = 10276; therefore

2 Kings 15:30 & 2 Kings 15:32 agree

2 Kings 15:30: & 2 Kings 15:32-33 seem to contradict each other in that 2 Kings 15:30 speaks of the twentieth year of Jotham while 2 Kings 15:33 reads, he (Jotham) reigned sixteen years in Jerusalem. We make sense of this apparent contradiction with the following verses:

2 Kings 15:³⁷In those days the LORD began to send against Judah Rezin the king of Syria, and Pekah the son of Remaliah. ³⁸And Jotham slept with his fathers, and was buried with his fathers in the city of David his father: and Ahaz his son reigned in his stead 

Jotham reigned for 16 years (according to 2 Kings 15:33). Then Pekah (and Rezin) invaded Judah and took away his power, though Jotham still hung around for another 4 years (according to 2 Kings 15:30).

(2 Kings 15:30 & 2 Kings 15:32) → (11p ∆ 11q) = 10276 →

11p = (10275 ∆ 10276) & 11q(20) = 11Q = (10295 ∆ 10296).

2 Kings 15:30 → Hoshea≡20p_I = 10295 as the 20th year of Jotham. Since Jotham wasn't around after his 20^th year, 11Q ≠ 10296 and the 20^th year in 2 Kings 15:30 refers to 11q(20). Therefore 11Q = 11q(20) = 10295 &

11q = 10276 & 11p = 10275

In this instance only does Jotham's reign define the year of another king. Because 2 Kings 15:30 refers to Jotham≡11q(20) = 11Q, we have further corroboration that unless the data directs us to point p, we know the verse refers to point q (as explained on November 17 under 2 Kings 13:9). The exceptions that did refer us to point p: 1 Kings 15:28 regarding Asa (see November 5); 2 Kings 1:16-17 regarding Jehoram (see November 12 un- der 2 Chronicles 21:20); 2 Kings 13:1 regarding Jehoash (see November 17); 2 Kings 15:23 regarding Pekahiah.

2 Kings 15:30 → 11q(20) = 10295 &

Jotham≡11q = (10295 – 19) = 10276

11q≡Jotham reigned for 20 years; from 10275 until 10295 (=11Q)

2 Kings 15:37 → Ahaz≡12q ≤ 10295

On November 24 we look at Hoshea and Hezekiah and Ahaz.

Hoshea, Hezekiah & Ahaz

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 22:

2 Kings 15:25, 27 → Pekah≡19p_I = 10275

2 Kings 15:37 → Jotham≡11Q = 10295

2 Kings 15:30 → Hoshea≡20p_I = 10295

(That in green is for those needing help in understanding the formal arguments in aqua.)

2 Kings 16:¹In the seventeenth year of Pekah the son of Remaliah Ahaz the son of Jotham king of Judah began to reign. ²Twenty years old was Ahaz when he began to reign, and reigned sixteen years in Jerusalem, and did not that which was right in the sight of the LORD his God, like David his father.

2 Kings 15:25, 27, 30, 37, 38 →

19p_I≡Pekah = 10275

2 Kings 16:1 → for some n,

Ahaz≡12p(n) = (19p_I(17) ∆ 19q_I(17)) = (10291 ∆ 10292)

2 Kings 16:2 → 12q(16) = 19Q

2 Kings 17:¹In the twelfth year of Ahaz king of Judah began Hoshea the son of Elah to reign in Samaria over Israel nine years →

2 Kings 17:1 →

Hoshea≡20p_I = Ahaz≡(12p(12) ∆ 12q(12)) &

2 Kings 15:30 → 20p_I = 10295; therefore

(12p(12) ∆ 12q(12)) = 10295 →

(12p ∆ 12q) = 10284. Since p(n) ± m = p(n ± m).

Because the data does not refer us to 12p we assume

12q = 10284 (see November 22 under 2 Kings 15:37)

2 Kings 16:1 → 12q(16) = 12Q = 10299

12q≡Ahaz reigned for 16 years: from 10283 until 10299 (= 12Q)

2 Kings 16:1-2 refers to 19q_I(17) to define 12p(n); i.e.,

12p(n) = 19q_I(17) = 10292 & n = 9. Also 12q(a) = 10296 → a = 13

Beginning in 10283 (= 12p) Ahaz ruled as a co-regent with his father. From 2 Kings 15:30 & 2 Kings 15:32-33 & 2 Kings 15:37 we know Ahaz ruled for 4 years (= 20 – 16) after Pekah invaded Judea and took Jotham captive (see November 22 under 2 Kings 15:37-38). From 2 Kings 15:30, 32 we know Jotham ≡11Q = 10295. So Israel invaded Judea in 10292. Therefore during his 9^th (= n^th) year, i.e., 12q(9) = 10292, this transition of power took place that favored Ahaz; and 2 Kings 16:1-2 → 19q_I(17) = 12p(n) = 10292. So 2 Kings 16:1-2 uses 19q_I(17) to define 12p(n) and corroborates the principle that if the data does not refer us to point p, then the verse in question refers to point q. In his (Ahaz) 12^th year, i.e., 12q(12) = 10295, Jotham died. Beginning with 10296 = 12q(13 (= a)), Ahaz ruled for 4 more years without any paternal input, whatever that might have been.

2 Kings 15:37-38 → Hoshea≡20p_I = 10295 & 2 Kings 17:1 →

20q_I(9) = 20Q_I = (10295 + 9) = 10304.

20q_I≡Hoshea reigned for 9 years: from 10295 until 10304 (=20Q_I).

2 Kings 17:⁶In the ninth year of Hoshea the king of Assyria took Samaria, and carried Israel away into Assyria, and placed them in Halah and in Habor by the river of Gozan, and in the cities of the Medes….

2 Kings 17:6 → 20q_I(9) = 20Q_I therefore 2 Kings 17:6 refers to 20q_I(9) to define 20Q_I. This corroborates the principle that if the data does not refer us to point p then the verse in question refers to point q.

That does it for ancient Israel and their 20 kings. We find the reason for their bitter end in 2 Kings 17:¹⁸…the LORD was very angry with Israel, and removed them out of his sight: there was none left but the tribe of Judah only.

2 Kings 18:¹ Now it came to pass in the third year of Hoshea son of Elah king of Israel, that Hezekiah the son of Ahaz king of Judah began to reign. ²Twenty and five years old was he when he began to reign; and he reigned twenty and nine years in Jerusalem. His mother's name also was Abi, the daughter of Zachariah. 

2 Kings 17:1 → Hoshea≡20p_I = 10295 & 2 Kings 18:1 →

Hezekiah≡13p = Hoshea≡(20p_I(3) ∆ 20q_I(3)) &

13p = (10297 ∆ 10298). Since p(n) ± m = p(n ± m).

Hezekiah≡13q(29) = 13Q.

2 Kings 18:⁹And it came to pass in the fourth year of king Hezekiah, which was the seventh year of Hoshea son of Elah king of Israel, that Shalmaneser king of Assyria came up against Samaria, and besieged it. ¹⁰And at the end of three years they took it: even in the sixth year of Hezekiah, that is the ninth year of Hoshea king of Israel, Samaria was taken. 

2 Kings 17:1 → Hoshea≡20Q_I(9) = 20Q_I = 10304 →

2 Kings 18:10→ (13p(6) ∆ 13q(6)) = 10304 →

(13p ∆ 13q) = 10299. Since p(n) ± m = p(n ± m) &

2 Kings 18:1 → 13p = (10297 ∆ 10298); therefore

13p = 10298; since if 13p = 10297 then 13q = 10298 which contradicts (13p ∆ 13q) = 10299. Therefore 13p ≠ 10297 and 13p = 10298.

2 Kings 18:10 → 13q(6) = 10304 → 13q = 10299; therefore 2 Kings 18:10 refers to 13q(6) to define Hoshea≡20Q_I

2 Kings 18:9 → 20q_I(7) = 13q(4)

2 Kings 15:30 → 20p_I = 10295

2 Kings 18:10 → 13p = 10298

(10295 + 7) = (10298 + 4) = 10302 therefore

2 Kings 18:9 uses 20q_I(7) & 13q(4) to define 10302.

2 Kings 18:1→ 20q_I(3) = 13p = 10298 therefore 2 Kings 18:1 uses 20q_I(3) to define 13p.

2 Kings 18:1 & 2 Kings 18:9 & 2 Kings 18:10 corroborate the principle that if the data does not refer us to point p then the verse in question re- fers to point q.

2 Kings 18:2 → 13q(29) = 13Q. Therefore 13Q = 10327

13q≡Hezekiah ruled for 29 years: from 10298 until 10327 (= 13Q)

On November 26 we look at Manasseh, Amon, Josiah, Jehoahaz & Jehoiakim.

Manasseh, Amon, Josiah, Jehoahaz & Jehoiakim

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 24: 13q≡Hezekiah ruled for 29 years: from 10298 until 10327 (= 13Q)

To get an idea of Hezekiah's situation we go to 2 Kings 20:

¹In those days was Hezekiah sick…And the prophet Isaiah…said unto him, Thus saith the LORD…thou shalt die, and not live. ² Then he (Hez- ekiah)… prayed unto the LORD…⁴And...the LORD came to him (Isa- iah), saying…, ⁵tell Hezekiah…I will heal thee…⁶And…add unto thy days fifteen years…⁷And Isaiah said, Take a lump of figs. And they took and laid it on the boil, and he recovered…^8…21And Hezekiah slept with his fathers: and Manasseh his son reigned in his stead (2 Kings 20). ¹Manasseh was twelve years old when he began to reign, and reigned fifty and five years in Jerusalem. And his mother's name was Hephzibah (2 Kings 21).

Hezekiah did not die an untimely death. Since 10327 – 15 = 10312 & 10312 – 14 = 10298 = 13p, he knew in his 14th year that he had 15 years to live. With 14p≡Manasseh at 12 years, 13q(15) < 14p < 13q(29), though most likely 14p falls closer to 13q(15) then 13q(29). The question is, "When?"

In this only instance Scripture doesn't give enough information to deter- mine Manasseh≡14p. So we have to approximate as best we can: in 13q(15) we can imagine Hezekiah kicking his heels over his pardon. But in about two years he begins to realize that just because God put him on hold temporarily doesn't mean his allotted time won't run out. So he fig- ures he'd better groom his son, Manasseh, to keep the seat warm on the throne. Give him another year to finally get around to it—you know how it goes—and we'll guess at 14p ≈ 13q(18). Technically we could be 11 years short, but that's so unlikely when Hezekiah knew his last year.

Using 14p ≈ 10316 (= 10298 (= 13p) + 18), we calculate the reigns of the five remaining kings. We then map our timeline onto the Gregorian calen- dar and make any necessary adjustments to 14p so that the last king, Zed- ekiah≡20Q = 587 B.C.E., when Babylon sacked Jerusalem.

2 Kings 21:1→14q(55) = 14Q = 10371.

14q≡Manasseh reigned for 55 years; from 10316 until 10371 (=14Q).

2 Kings 21:¹⁸And Manasseh slept with his fathers, and was buried…and Amon his son reigned in his stead. ¹⁹Amon was twenty and two years old when he began to reign, and he reigned two years in Jerusalem… ²³And the servants of Amon conspired against him, and slew the king in his own house. ²⁴And the people of the land slew all them that had conspired against king Amon; and the people of the land made Josiah his son king in his stead.

Because Amon reigned only two years we assume

14Q = 15p≡Amon = 10371 & 15q(2) = 15Q.

15Q = 10373 = Josiah≡16p

15q≡Amon reigned for 2 years; from 10371 until 10373 (=15Q).

2 Kings 22:¹Josiah was eight years old when he began to reign, and he reigned thirty and one years in Jerusalem...

Josiah≡16p = 10373

Josiah≡16q(31) = (10373 + 31) = 10404 = 16Q.

16q≡Josiah reigned for 31 years; from 10373 until 10404 (=16Q).

2 Kings 23:²⁹In his (i.e., Josiah) days Pharaohnechoh king of Egypt went up against the king of Assyria to the river Euphrates: and king Josiah went against him; and he slew him (Josiah) at Megiddo, when he had seen him. ³⁰And his servants carried him in a chariot dead from Megiddo, and brought him to Jerusalem, and buried him in his own sepulchre. And the people of the land took Jehoahaz the son of Josiah, and anointed him, and made him king in his father's stead. ³¹Jehoahaz was twenty and three years old when he began to reign; and he reigned three months in Jerusalem. 

Since Jehoahaz reigned only 3 months, and Josiah was killed unex- pectedly, we assume, Josiah≡16q(31) = 16Q = Jehoahaz≡17p.

17p≡Jehoahaz reigned for 3 months; from 10404 until 10404 (= 17Q).

2 Kings 23:³³And Pharaohnechoh put him (Jehoahaz) in bands at Rib- lah…³⁴And Pharaohnechoh made Eliakim the son of Josiah king in the room of Josiah his father, and turned his name to Jehoiakim, and took Jehoahaz away: and he came to Egypt, and died there.

Note that Pharaohnechoh took Jehoahaz≡17q(.25) = 17Q = 10404 off the throne and set up Eliakim, a.k.a. Jehoiakim≡18p = 10404. Egypt now controls Jerusalem and it's the beginning of the end.

2 Kings 23:³⁶Jehoiakim was twenty and five years old when he began to reign; and he reigned eleven years in Jerusalem →

Jehoiakim≡18q(11) = (10404 + 11) = 10415 = 18Q.

18q≡Jehoiakim reigned for 11 years; from 10404 until 10415 (=18Q).

We complete the reigns of the Hebrew kings on November 29.

Jehoiachin, Zedekiah & Nebuchadnezzar

On October 27 we began with the Hebrew Kings and on October 29 and November 1 we completed the algebra for their calculations. On Novem- ber 3 we began with the first three kings: Rehoboam Jerobaom Abijah. In continuing we need to know the following from November 26:

2 Kings 23:33, 36 →

18q≡Jehoiakim reigned for 11 years; from 10404 until 10415 (=18Q).

On November 26 we left off with Egypt ruling over Judah. But then Ne- buchadnezzar, king of Babylon, defeated Egypt and now he taxes Jeru- salem and dictates who rules there.

Jeremiah 25:¹The word that came to Jeremiah concerning all the people of Judah in the fourth year of Jehoiakim the son of Josiah king of Judah, that was the first year of Nebuchadrezzar king of Babylon.

So as not to confuse Nebuchadnezzar (= Nebuchadrezzar) with a Hebrew king, we denote him as p_n and q_n.

2 Kings 23:33, 36 → 18p = 10404 & Jeremiah 25:1 →

p_n = (18p(4) ∆ 18q(4)) = ((10404 – 1) + 4) ∆ (10404 + 4)

p_n = (10407 ∆ 10408)

2 Chronicles 36:⁶Against him (18q≡Jehoiakim) came up Nebuchadnezzar king of Babylon, and bound him in fetters, to carry him to Babylon. ⁸Now the rest of the acts of Jehoiakim… they are written in the book of the kings of Israel and Judah: and Jehoiachin his son reigned in his stead & 

2 Kings 23;33, 36 → Jehoiachin≡19p = 18q(11) = 18Q

2 Chronicles 36:⁹Jehoiachin was eight years old when he began to reign, and he reigned three months and ten days in Jerusalem: and he did that which was evil in the sight of the LORD. ¹⁰And when the year was ex- pired (Nisan 1 in March or April), king Nebuchadnezzar sent, and brought him to Babylon, with the goodly vessels of the house of the LORD, and made Zedekiah his brother king over Judah and Jerusalem.

2 Chronicles 36:10 → 20q≡Zedekiah = 19Q &
for all n = 1, 2, 3…11, Zedekiah≡20p(n) = Zedekiah≡20q(n)

Nebuchadnezzar brought Jehoiachin to Babylon when the (Jewish) year was expired. He then put Zedekiah≡20p (a.k.a. Mattaniah) on the throne on or before Nisan 1, the first month of the Jewish calendar (see Exodus 12:1-2). Because this lunar year begins in March or April, Zedekiah≡20p would have sat on the throne and begun to rein officially in the same Gre-gorian year; i.e., 20p = 20q. Until now we didn't have to worry about a king expiring between January 1 and Nisan 1 because the killing season usually ran through the summer, when warmer weather made assassina- tions and dying in battle more fun. But in this one exception we do not have 20q – 1 = 20p; rather for all n = 1, 2, 3…11, 20p(n) = 20q(n).

2 Kings 24:⁸Jehoiachin was eighteen years old when he began to reign, and he reigned in Jerusalem three months. And his mother's name was Nehushta, the daughter of Elnathan of Jerusalem & 2 Chronicles 36:9 →

Jehoiachin≡19q(10.27) = 19Q.

2 Chronicles 36:9 → Jehoiachin≡19p (a.k.a. Jechonias and Coniah) was 8 years old when he began to reign; 2 Kings 24:8 → Jehoiachin≡19p(n) was 18 years old when he began to reign. Therefore 19p reigned as co-regent with his father, Jehoiakim≡18p, for ten years (18 – 8); then when 19p was 18 years old, Nebuchadnezzar≡p_n brought his father, 18p, to Babylon (2 Chronicles 36:6), leaving 19p to reign by himself for t = .27 (= 3 months and 10 days of 2 Chronicles 36:9) years.

2 Kings 23:33, 36 → 18q(11) = 18Q.

2 Chronicles 36:8-10, & 2 Kings 24:8 →

18q(11) + t = 19q(10) →

18q + t = (19q – 1) = 19p

2 Kings 23:36 → 18q + t = 10405 + t →

19p = 10405 + t &

19q(10) = 19Q = (10405 + 10 + t) = 10415 + .27.

19p≡Jehoiachin reigned for 10.27 years; from 10405 until 10415 or 10416 (= 19Q).

2 Kings 24:¹²And Jehoiachin the king of Judah went out to the king of Babylon, he, and his mother, and his servants, and his princes, and his officers: and the king of Babylon took him in the eighth year of his reign →

(p_n(8) ∆ q_n(8)) = 19q(10.27) & 2 Kings 24:8 →

(p_n(8) ∆ q_n(8)) = (10415 ∆ 10416).

2 Kings 24:¹⁷And the king of Babylon made Mattaniah his father's brother king in his stead, and changed his name to Zedekiah. ¹⁸Zedekiah was twenty and one years old when he began to reign, and he reigned eleven years in Jerusalem. And his mother's name was Hamutal, the daughter of Jeremiah of Libnah

2 Kings 24:18 → 20q(11) = 20Q.

Jeremiah 32:¹The word that came to Jeremiah from the LORD in the tenth year of Zedekiah king of Judah, which was the eighteenth year of Nebuchadrezzar →

20q(10) = (p_n(18) ∆ q_n(18))

After overdosing on stupid pills, Zedekiah rebels against Nebuchadnezzar. So after slaying his sons, the king of Babylon puts out Zedekiah's eyes (Jeremiah 39:6-7) and brings him to Babylon to die. Nebuchadnezzar had never intended to destroy Jerusalem, just govern it. But now, thanks to Zedekiah, he has a change of heart:

Jeremiah 52:¹²Now in the fifth month, in the tenth day of the month, which was the nineteenth year of Nebuchadrezzar king of Babylon, came Nebuzaradan, captain of the guard, which served the king of Babylon, into Jerusalem, ¹³And burned the house of the LORD, and the king's house; and all the houses of Jerusalem, and all the houses of the great men, burned he with fire: ¹⁴And all the army of the Chaldeans, that were with the captain of the guard, brake down all the walls of Jerusalem round about. ¹⁵Then Nebuzaradan the captain of the guard carried away captive certain of the poor of the people, and the residue of the people that re- mained in the city, and those that fell away, that fell to the king of Baby- lon, and the rest of the multitude.

2 Kings 24:17-18 & Jeremiah 52:12 →

20q(11) = 20Q = Nebuchadnezzar≡(p_n(19) ∆ q_n(19)).

Jeremiah 32:1→ 20q(10) = (p_n(18) ∆ q_n(18)) & 

if p_n(18) = 20q(10) then

p_n(18) + 1 = 20q(10) + 1 → p_n(19) = q_n(18) = 20q(11) = 20Q so Jeremiah 32:1 agrees with Jeremiah 52:12; i.e., Nebuchadnezzar's 18^th (= 20Q) year (according to Jeremiah 32:1) is also his 19^th (= 20Q) year (according to Jeremiah 52:12).

Also p_n(19) = 20q(11) → p_n(9) = 20q.

Jeremiah 52:²⁸This is the people whom Nebuchadrezzar carried away captive: in the seventh year three thousand Jews and three and twenty: ²⁹In the eighteenth year of Nebuchadrezzar he carried away captive from Jerusalem eight hundred thirty and two persons.

Jeremiah 52:28 → (p_n(7) ∆ q_n(7)): Nebuchadrezzar carried away captive: three thousand Jews…

2 Kings 24:8, 12 →

19q(10.27) = (p_n(8) ∆ q_n(8)) = (10415 ∆ 10416).

Therefore (Jeremiah 52:28 & 2 Kings 24:12) →

(p_n(8) = q_n(7)) = (10415 ∆ 10416) →

p_n = ((10415 – 7) ∆ (10416 – 7)) →

p_n = (10408 ∆ 10409)

Jeremiah 25:1 → p_n = (10407 ∆ 10408); therefore.

p_n = 10408

Jeremiah 52:12-15 → p_n(9) = 20q(= 20p) & p_n = 10408 →

p_n(9) = (10408 – 1) + 9 = 10416 →

20p(11) = (10416 – 1) + 11) = 10426.

20q≡Zedekiah reigned for 11 years: from 10416 until 10426 (= 20Q).

Jeremiah 52:29 → In (p_n(18) ∆ q_n(18)) Nebuchadnezzarcarried away captive from Jerusalem eight hundred thirty and two persons &

Jeremiah 52:12-15 → p_n(19) = 20q(11) = 20Q

Therefore p_n(19) = q_n(18) = 20q(11) = 20Q when Nebuchadnezzar carried away captive from Jerusalem eight hundred thirty and two persons.

Jeremiah 52:28-29 agrees with Jeremiah 52:12-15.

This completes the reigns of the Hebrew kings. On December 1 we map our timeline onto the Gregorian calendar.