Show that $(x_1\times\dots\times x_k)\times x_{k+1}=x_1\times\dots\times x_{k+1}$

Question

Let $a\times b\times c$ denote $a\times(b\times c)$.

Given $(a\times b)\times c=a\times(b\times c)$, how do you prove

$$(x_1\times\dots\times x_k)\times x_{k+1}=x_1\times\dots\times x_{k+1}$$

?

$a$, $b$, $c$, and $x_i$ for integers $i$ are members of a group and the operator that the group has is $\times$. However, I need to prove it from only the facts I gave before, and no other group axioms that exist.

@MartinBrandenburg: To my knowledge, it should not matter what they are. The only information given is the one in my post. But let's assume a bit. — user162425, Jul 09 '14 at 08:49
You should define the right hand side too, if you want to be formal. By recursion, I imagine. Then the proof by induction should be clear. — Henno Brandsma, Jul 09 '14 at 09:05

score 1 · Answer 1 · answered Jul 09 '14 at 09:02

I assume that also $x_1\times x_2\times \ldots \times x_{k-1}\times x_k$ means $x_1\times (x_2\times (\ldots \times (x_{k-1}\times x_k)\ldots))$. Then the proof goes by induction with the induction step being $$ \begin{align}(x_1\times x_2\times \cdots\times x_k)\times x_{k+1} &=(x_1\times (x_2\times \ldots\times x_k))\times x_{k+1}\\ &=x_1\times ((x_2\times \ldots\times x_k)\times x_{k+1})\\ &=x_1\times (x_2\times \ldots\times x_k\times x_{k+1})\\ &=x_1\times x_2\times \ldots\times x_k\times x_{k+1}\\ \end{align}$$ Of course, to make this rigourous, you'd have to get rid of "$\ldots$" even in the notation.

Follow up question: How would that be done? (getting rid of $\dots$ between the closing parentheses) — user162425, Jul 09 '14 at 09:04

Show that $(x_1\times\dots\times x_k)\times x_{k+1}=x_1\times\dots\times x_{k+1}$

1 Answers1