## Denominated Mischief

I’ve finally got around to reading one of my Christmas presents, Alfred S Posamentier and Ingmar Lehman’s Magnificent Mistakes in Mathematics, which is about ways that mathematical reasoning can be led astray. A lot, at least in the early pages, is about the ways a calculation can be fowled by a bit of carelessness, especially things like dividing by zero, which seems like such an obvious mistake that who could make it once they’ve passed Algebra II?

They got to a most neat little erroneous calculation, though, and I wanted to share it since the flaw is not immediately obvious although the absurdity of the conclusion drives you to look for it. We begin with a straightforward problem that I think of as Algebra I-grade, though I admit my memories of taking Algebra I are pretty vague these days, so maybe I missed the target grade level by a year or two.

Multiply that 4 on the right-hand side by 1 — in this case, by — and combine that into the numerator:

Expand that parentheses and simplify the numerator on the right-hand side:

Since the fractions are equal, and the numerators are equal, therefore their denominators must be equal. Thus, and therefore, 11 = 7.

Did you spot where the card got palmed there?

## Little Monster Girl 11:13 pm

onMonday, 2 February, 2015 Permalink |You didn’t do anything to the left side of the equation?

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## Joseph Nebus 10:20 am

onTuesday, 3 February, 2015 Permalink |It’s true nothing’s done on the left-hand side, but that isn’t by itself an error. If we start from the assumption that the original equation is true we can manipulate one side, or the other, or both, into a form that’s more convenient without changing whether or not the whole equation is true. The catch is that somewhere in this is a manipulation that doesn’t preserve the truth of the whole thing.

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## howardat58 11:30 pm

onMonday, 2 February, 2015 Permalink |Formally, cross multiply is in order. Of course, they don’t call it that these days.

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## Joseph Nebus 10:21 am

onTuesday, 3 February, 2015 Permalink |Cross-multiplying ought to give a fair shot at avoiding the error, yeah. But I couldn’t blame someone for seeing an equation of the form a/b = a/d and going right to b = d directly.

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## ivasallay 7:41 am

onTuesday, 3 February, 2015 Permalink |Where did (3x – 30)/ (11 – x) = (x + 2)/(x – 7) – 4 come from? It certainly isn’t true for all x.

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## Joseph Nebus 10:23 am

onTuesday, 3 February, 2015 Permalink |Well, that’s just a problem to be solved, to find values of x which make it true. It’s just that along the way to finding those x’s, we end up with a conclusion that 11 equals 7.

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## elkement 7:18 pm

onTuesday, 3 February, 2015 Permalink |I think the trick is to keep in mind that when the numerator is zero then it does not matter if both denominators are different (as long as they are not equal to zero as well).

So if x is equal to 10 the equation is true as both sides are equal to zero although the denominators are 1 and -3, respectively.

The short version is: You must not divide both sides of an equation of zero.

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## Joseph Nebus 11:31 pm

onWednesday, 4 February, 2015 Permalink |That’s it exactly, and I’m delighted by the problem since it

isone in which the ever-forbidden division by zero is made nicely non-obvious.LikeLiked by 1 person

## How February 2015 Treated My Mathematics Blog | nebusresearch 9:52 pm

onSunday, 1 March, 2015 Permalink |[…] Denominated Mischief, in which a bit of arithmetic manipulation proves that 7 equals 11. […]

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