Let ε < 0.


Mathematical limericks, vol. 3

Filed under: Harmonic analysis — Travis @

If you’re mathematically inclined, you’ve probably seen several proofs that 2=1. On of the more popular ones is the following:

This can be expressed as a three-stanza limerick:

If a=b (so I say)
And we multiply both sides by a
Then we’ll see that a-squared
When with ab compared
Are the same. Remove b-squared. Okay?

Both sides we will factorize. See?
Now each side contains a minus b.
We’ll divide through by a
Minus b and olé
a+b=b. Oh whoopee!

But since I said a=b,
b+b=b you’ll agree?
So if b = 1
Then this sum I have done
Proves that 2 = 1.


  1. There is an error is this proof. Going from lines 4 to 5:
    (a+b)(a-b) = b(a-b)
    a+b = b
    This is not a valid step. You are actually dividing both sides by (a-b) to cancel out (a-b). But a=b; therefore (a-b) = 0. Since any number real or complex divided by zero gives the empty set, then this step is invalid. In other words you cannot divide by zero. Proof disproved!

    Comment by John — 03.29.10 @

  2. since a=b then a-b in line 4 is equal to 0. from line 4 to 5 both sides were divided by a-b..but a-b is zero and dividing by zero is undefined…

    Comment by kookie — 04.30.10 @

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