I understand how induction works but I am stuck on how I should approach this problem. I know I could start with the base case, but I'm not sure if my approach would be a solid proof. Here is the question:

Prove that for all natural numbers n and all real numbers $x_{1}$, ..., $x_{n}$

$$ \big| \prod\limits_{i=1}^n{x_{i}} \big| = \prod\limits_{i=1}^n|{x_{i}}| $$

Now I could start with the usual base case, but can I assume that this is true?

n = 1: $$ \big| \prod\limits_{i=1}^1{x_{i}} \big| = |{x_{i}}| $$ and $$ \prod\limits_{i=1}^1|{x_{i}}| = |{x_{i}}| $$

Setting me in the right direction would be much appreciated thanks.


Define recursively: $\prod_{i=1}^{0}x_i:=1$ and $\prod_{i=1}^{n+1}x_i: =x_{n+1} \cdot\prod_{i=1}^{n}x_i$ for $n\ge0$.

The base case is trivial. Now suppose we have already proven for $n\ge 0$. Then

$$\bigg| \prod_{i=1}^{n+1}x_i \bigg|=\bigg|x_{n+1} \cdot \prod_{i=1}^{n}x_i \bigg|=|x_{n+1}| \cdot \bigg| \prod_{i=1}^{n}x_i \bigg|$$

And by inductive hypothesis we know that $\bigg| \prod_{i=1}^{n}x_i \bigg|= \prod_{i=1}^{n}|x_i| $. Thus

$$\bigg| \prod_{i=1}^{n+1}x_i \bigg|=|x_{n+1}|\prod_{i=1}^{n}|x_i|=\prod_{i=1}^{n+1}|x_i| $$

You do have to prove $|a \cdot b|= |a||b|$ for $a,b$ real numbers first.

  • $\begingroup$ No. Because there is no natural number such that $ 1 \le i \le 0$. So in that undefined case we select the value of $1$. Thus for instance we have the identities: $$\prod_{i=1}^0x_i=1,\,\; \prod_{i=1}^{1}x_i=x_1\cdot\prod_{i=1}^0x_i=x_1\cdot1=x_1$$ and so on. $\endgroup$ Feb 26 '14 at 6:12
  • $\begingroup$ As for "define recursively", did you arbitrarily choose $x_{i}$ to be 1? Thanks a lot $\endgroup$
    – Kevin
    Feb 26 '14 at 6:13
  • $\begingroup$ en.wikipedia.org/wiki/Recursive_definition $\endgroup$ Feb 26 '14 at 6:18
  • $\begingroup$ We define the finite product by setting $\prod _{i=1}^{0}x_i=1$, and $\prod _{i=1}^{n+1}x_i=x_{n+1} \cdot \prod _{i=1}^{n}x_i$. So you can check that $\prod _{i=1}^{2}x_i=x_2 \cdot x_1$, $\prod _{i=1}^{3}x_i=x_3 \cdot x_2 \cdot x_1$. Informally $$\prod _{i=1}^{n}x_i=x_n \cdot x_{n-1}\ldots x_1$$ I'm not choose arbitrary $x_i=1$ $\endgroup$ Feb 26 '14 at 6:28
  • $\begingroup$ Ah, I understand. I think my confusion arose because I had thought that 0 was not a natural number (debated of course), but in the case where I assume n $\neq$ 0, then there would be no need for the recursive definition right? $\endgroup$
    – Kevin
    Feb 26 '14 at 6:45

The product of one number is just the number, isn't it?

  • $\begingroup$ Yes, I think that the base case is correct, and then I could substitute for n = k, but then to use the inductive hypothesis for n = k + 1 seems too trivially simple. $\endgroup$
    – Kevin
    Feb 26 '14 at 5:33
  • $\begingroup$ Maybe it is a little simple. Another way you can think of it is, just prove this for when $n = 2$, since every other $n > 2$ can be reduced to the case of $n = 2$. $\endgroup$ Feb 26 '14 at 5:37
  • $\begingroup$ By that do mean let $x_{i}$ be say when n = k, therefore k + 1 would be as if n = 2? $\endgroup$
    – Kevin
    Feb 26 '14 at 5:48
  • $\begingroup$ I meant to say that the problem of multiplying a sequence of numbers can be reduced to the multiplication of 2 numbers at a time. For instance, $|x_1| \cdot |x_2| \cdot |x_3| = |x_1x_2| \cdot |x_3| = |x_1x_2x_3|$ $\endgroup$ Feb 26 '14 at 5:52
  • $\begingroup$ Ah that makes a sense, but would that still be considered inductive? $\endgroup$
    – Kevin
    Feb 26 '14 at 6:14

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