0

Say we are trying to prove $$ 0\cdot n = 0 $$ By mathematical induction, we start with a base case of n = 1 $$ 0\cdot 1 = 0 $$ So now we assume our original formula is true, and try to prove a case for $n+1$. $$ 0\cdot (n+1) = 0\cdot n + 0\cdot 1 = 0 $$ Why can't we take the limit as n approaches infinity?

This would tell us that infinity times zero is equal to zero.

Aaron Maroja
  • 17,869
Bagel
  • 1
  • 1
  • 4
    If this did work, you could also use it to prove that infinity was a finite number: 0 is a finite number, and if $n$ is finite, then $n+1$ is finite, therefore… – MJD Feb 17 '15 at 19:45
  • You can indeed take limit as $n\to\infty$. You get the valid equation $\lim_{n\to\infty}(0(n))=0$. The step that is problematic is then to say that $\lim_{n\to\infty}0(n)=\lim_{n\to\infty}0\lim_{n\to\infty}n$. – Tom Feb 17 '15 at 19:52
  • Related: http://math.stackexchange.com/q/98093/ – Jonas Meyer Feb 18 '15 at 02:58

1 Answers1

6

Mathematical induction in the way you used it allows you to prove that a statement is true for all natural numbers (positive integers). Infinity is not a natural number, so your proof doesn't apply to infinity.

What you've shown is that for all $n \in \mathbb{N}$, $$0 \cdot n = 0$$

Zubin Mukerjee
  • 18,311