I'm a sucker for those math problems that go around Facebook.

x = 6 \div 2(1 + 2) \\ 
x = \frac{6}{2(1 + 2)} \\
x = \frac{6}{2 + 4} \\
x = \frac{6}{6} \\
x = 1 \\
y = 6 \div 2(1 + 2) \\
y = \frac{6}{2(1 + 2)} \\
y = \frac{2 \cdot 3}{2(1 + 2)} \\
y = \frac{2 (1 + 2)}{2(1 + 2)} \\
y = 1

Added what appear to be the original form of the problem.

x = 6 \div 2(1 + 2) \\
x = \frac{6}{2(1 + 2)} \\
x = \frac{6}{2 + 4} \\
x = \frac{6}{6} \\
x = 1 \\
y = 6 \div 2(1 + 2) \\
y = \frac{6}{2(1 + 2)} \\
y = \frac{2 \cdot 3}{2(1 + 2)} \\
y = \frac{2 (1 + 2)}{2(1 + 2)} \\
y = 1 \\
z = 6/2 \space \space (1 + 2) \\
z = \frac{6}{2}(1 + 2) \\
z = \frac{\tfrac{6}{2}}{\tfrac{12}{4}} \\
z = \frac{3}{3} \\
z = 1
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The internet claims that the correct answer is actually 9. In this situation it's one of the reasons I hate the "division" operator when writing things out mathematically. Algebra uses the fractional notation for a reason: It's unambiguous.

Interestingly the form it was presented in best translates to way a line in a program might write it. It's a bit funny to me that I see the algebraic translation first before I see the pure programmatical version.

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x = 6 \div 2(1 + 2) \\
x = 6 \div 2 \cdot (1 + 2) \\
x = \frac{6}{2} \cdot (1 + 2) \\
x = 3 \cdot (1 + 2) \\
x = 3 \cdot 3 \\
x = 9
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youtube.com/watch?v=URcUvFIUIh

This video claims that this is a "historical usage" that was deprecated in 1917. As a matter of historical fact I have no clue if this is true. I wasn't alive in 1917 though so I suspect this notation was alive and well up until at least 2001.

The example they give is pretty good and exactly explains how I think about it.

6 \div 2(1 + 2) \\
x \div y(j + k) \\
\frac{x}{y(j + k)} \Leftrightarrow x/(y(j+k))
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@gnu_lorien Without watching the video, I suspect that what happened is some guy in 1917 pointed out the problem, but he wasn't the King of Math so people kept doing it.

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