Erik Anderson’s Blog Debut at Study Hut Tutoring

Here’s a string of algebra I came across this weekend, see if you can find the error:

1. a = b
2. (a^2) = ab
3. (a^2) + (a^2) = (a^2) + ab
4. 2(a^2) = (a^2) + ab
5. 2(a^2) – 2ab = (a^2) + ab – 2ab
6. 2(a^2) – 2ab = (a^2) – ab
7. 2((a^2) – ab) = 1((a^2) – ab)
8. 2 = 1

What happened? In which step does the mistake lie? If a = 2 and b = 1 that a does not equal b.

But there didn’t appear to be any errors… so are all numbers really arbitrary and meaningless? Does 2 really equal 1?

No, it doesn’t. Where is the error?

If you said step one, you’re wrong. a does equal b, we go about the whole problem on that assumption. Step two is correct too and it’s all fine up until about step five, when we subtract 2ab. On step six, upon closer examination, both sides are now equal to zero.
a squared minus ab equals zero. But this is not where the mistake lies. If you said step seven, you’re correct. By dividing both sides by ((a^2) – ab) we were effectively dividing by zero.

In algebra, dividing by zero never goes well. By the time we get to calculus, we begin to see that almost dividing by zero results in very large numbers that approach infinity. That is limits.

So what happened by step 8 is that when we divide two by a very small number (or zero) it results in the same indeterminable, infinite quantity that results in dividing one by a very small number (or zero).

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Date posted: Thursday, December 6th, 2007 1:02 AM | Under category: Tutors
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1 Comment

  1. Mike Stone said »

    Erik,

    Great post! I think my head is going to explode. Keep up the good postings and tell Rob that he should have posts on this blog at least 3-5x a week.

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