I've noted that NUMALGEBRA (⇒), or ALGEBRA (⇐), in general, cannot be adequately defined without the notion of SETS. The SOLUTION to an ALGEBRAIC PROBLEM -- whatever the ALGEBRA (⇐) -- is a SET OF MEMBERS, each ELEMENT SATISFYING THE CONDITION OF THE PROBLEM. But I don't know where you will find this today in any text, article, paper, etc.

Curiouser and curioser (as Alice said), you find it in ancient Egyptian hieroglypics on clay or papyrus. The Egyptian word was "aha", meaning "heap". So, the "backwards" number or numbers sought by Egyptian priests (to understand the future from the past) was a member of a heap or a set.

We know this from the Rhind Papyrus (circa 1650 B.C.) in The British Museum in London. Here is a translation of one of its "aha" problems:
"Problem 24: A quantity and its 1/7 added together become 19. What is the quantity?

Assume 7. 1 1/7 of 7 is 8. As many times as 8 must be multiplied to give 19, so many times 7 must be multiplied to give the required number."

This is "rhetorical algebra" -- the first stage of algebra (⇐) formulated in words. In the second stage, "syncopated algebra", Latin words were abbreviated, such as eq for "equus", Latin for "equal.

In third stage "symbolic algebra", we can write:
x + x/7 = 19 → 8x/7 = 19 → 8x = 7 · 19 = 133 → 8x = 133 → x = 133/8. Proof: 133/8 + 133/(7 · 8) = 133/8 + 19/8 = 152/8 = 19.

Later on, Chinese mathematicians were able to solve problems in two simultaneous algebraic () equations in two unknowns, such as the following (in rhetorical algebra) from the Han period (circa 200 B.Bc.):
"One pint of good wine costs 50 gold pieces, while one pint of poor wine costs 10. Two pints of wine are bought for 30 gold pieces. How much of each kind of wine was bought?"

Symbolically, let x denote fraction of good wine, y, the fraction of poor wine. Then we know that x + y = 2. Using the respective prices, we construct another equation: 50x + 10y = 30. We now have two INDEPENDENT EQUATIONS (one is not MULTIPLE of OTHER) in TWO UNKNOWNS, x, y, so should be able to solve for them.

x + y = 2 → y = 2 — x. SUBSTITUTING this in the 2nd EQUATION, we find: 50 x + 10 (2 — x) = 30 → 50 x + 20 — 10 x = 30 → 40 x + 20 = 30 → 40 x = 30 — 20 → 40 x = 10 → x = 10/40 → x = 1/4. SUBSTITUTING this in the MODIFIED 1st EQUATION, we find: y = 2 — 1/4 ⇒ y = 7/4. Checking: x + y = 2 ⇒ 1/4 + 7/4 = 8/4 = 2. Also, 50(1/4) + 10(7/4) = 30 → 50/4 + 70/4 = 30 ⇒ 120/4 = 30. CHECKS!

However, we apparently do not know if the ancient Chinese had a word like the ancient Egyptian "aha" ("heap"), which RECOGNIZE THAT THE UNKNOWN IS A SET, even if A SINGULAR SET. (Perhaps Chinese scholars can enquire as to this possible usage.)

But we know that "they had it correct, Once Upon a Time". And mathematicians and math teachers went astray from this Egyptian espousal.

Go! and syn no more!

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