The teacher gives Johnny and Mary’s class a routine problem involving rational numbers.
The Problem: “A goblet contains 100 mL of water (H_{2}O) and a bottle contains 100 mL of red wine. I pour 10 mL from the bottle into the goblet and then 10 mL from the goblet back into the bottle. How much wine is now in the goblet and how much water is now in the wine?”
Introduction: If you like wine, you may be wondering who would do that sort of thing. On the other hand, if you dislike wine, you could be thinking that textbooks have no business referring to alcoholic beverages. However, these and any number of other nonmathematical concerns will not solve the problem. Sadly, if you are the sort of person who is easily sidetracked by this sort of speculation you are unlikely to appreciate the usually contrived world of school mathematics.
Instead, the following rather technical suppositions are more useful:
· Volumes of water and wine are strictly additive. The total volume always remains 200 mL.
· Red wine cannot change into water and vice versa.
· On contact, red wine and water mix completely and cannot be unmixed. Thus, they must always be transferred proportionally.
· milliLitres, mL, of water or wine may be subdivided into microLitres, μL, and they in turn may be subdivided into nanoLitres, nL, and so on indefinitely…
A Note about Notation: The notation (W, R) will be used throughout. For example (90, 9) means 90 mL of water and 9 mL of red wine.
A First Attempt: The teacher asks, “What is wrong with the proposed solution, below?”
Goblet

10 mL spoon

Bottle
 
Initial condition

(100, 0)

(0, 0)

(0, 100)
 
During the 1st transfer

(100, 0)

(0, 10)

(0, 90)
 
After 1st transfer

(100, 10)

(0, 0)

(0, 90)
 
During the 2^{nd} transfer

(90, 10)

(10, 0)

(0, 90)

x

Mary explains that the second transfer is “illegal” as the ratio 90 : 10 is obviously not the same as 10 : 0. The first ratio simplifies to 9 : 1. The second ‘simplifies’ to 1 : 0 which some brave souls might perhaps rewrite as ∞ : 1. (As a practical matter, from the 100 : 10 mixture, how is it that we only select 10 parts of water to be transferred?)
A Second Attempt: Johnny suggests the following table, instead.
Goblet

10 mL spoon

Bottle
 
Initial condition

(100, 0)

(0, 0)

(0, 100)
 
During the 1st transfer

(100, 0)

(0, 10)

(0, 90)
 
After 1st transfer

(100, 10)

(0, 0)

(0, 90)
 
During the 2^{nd }transfer

(90, 9)

(10, 1)

(0, 90)

x

After 2nd transfer

(90, 9)

(0, 0)

(10, 91)

The ratio for the second transfer is correct since 90 : 10 <=> 10 : 1. However, the teacher reminds Johnny that he has fallen afoul of one of the “givens” of the word problem, pointing at: “and then 10 mL from the goblet back into the bottle.” With the hint, Johnny 'remembers' that the 2nd transfer was a total of 11 ml.
The Proposed Solution: Mary draws up the following table.
She exclaims, “Look, there is just as much water in the wine as there is wine in the water!”
She exclaims, “Look, there is just as much water in the wine as there is wine in the water!”
Is the solution correct?: While the solution is very elegant, Johnny wants to argue that it is not absolutely correct. In Chemistry class, he has been told there is a smallest piece of water, termed a molecule. Therefore, he argues we cannot write ‘90.909090 dot dot dot.’ Instead, at some point the decimal must truncate. It could be after 20 decimal points or 53 or some other number, but thereafter the pattern would be broken.
“Equivalently”, Johnny says, “We can divide 99 by 11 but we cannot divide 100 by 11. Mary can ‘smooth over’ that difficulty by converting 100 mL to 100 000 μL. Nevertheless, as that too will still not divide by 11, she must make a further change to 100 000 000 nL. But neither will that divide by 11. In typically human fashion she simply made the problem recede, not vanish. Eventually we must reach a smallest volume that contains a single molecule of water, forcing us to stop the process of ‘converting between units’.”
Are any of the solutions reasonable?: Mary appeals to the teacher, who confirms that for the purpose of IGCSE Mathematics, her answer is correct, as it is mathematically more sophisticated. But he reminds her that it is acceptable to round answers to 3 d.p., thus, (90.9, 9.09) and (9.09, 90.9).
The teacher further explains that Mathematics is an idealization of reality. In every problem, there is always a translation (or idealization) process from reality to mathematics, in which simplifying assumptions are made. If we do not simplify enough the problem may be insoluble. However, if we oversimplify, the conclusion we draw may be invalid. For IGCSE Mathematics these concerns are generally pushed to the background as they belong more to a Philosophy or to a “Theory of Knowledge” course and our focus is generally on ensuring students possess accurate and sophisticated methods of calculation.
Footnotes:
1) In the kitchen (but not on the exam page), if I transfer the water and the wine by a teaspoon, I am exceedingly unlikely to be able to distinguish between 10 mL and 11 mL. Thus, for many practical purposes Johnny’s attempted solution is good enough.
2) Johnny has continued to calculate and tells the class that 1 mL of water is 30 000 000 000 000 000 000 000 molecules. (Or near enough.) Consequently, a nanolitre is still an amazing 30 000 000 000 000 molecules. He now agrees that in leaving aside the dot dot dot, you can still conclude that the ratio of water to wine in the goblet is the same as the ratio of wine to water in the bottle.
3) A chemist who reads this blog may want to dispute the additivity of volumes of water and wine. (S)he will say 100 ml of water mixed with 100 ml of red wine actually makes less than 200 ml of solution! (But that is a story for another day).