Algebra is symbolic manipulation though painstaking or conscientious :-)

Of course, I have oversimplified the meaning of algebra. 🙂

Here is an example. Let me know what you think. (Reference: Algebra 3rd Edition by Serge Lang).

Let G be a commutative monoid, and $x_{1}, x_{2}, \ldots, x_{n}$ be elements of G. Let $\psi$ be a bijection of the set of integers $(1,2, \ldots, n)$ onto itself. Then, $\prod_{v=1}^{n}x_{\psi(v)} = \prod_{v=1}^{n}x_{v}$

Proof by mathematical induction:

PS: if one gets scared by the above notation, one can expand it and see its meaning. Try that.

It is clearly true for $n=1$. We assume it for $n=1$. Let k be an integer such that $\psi(k)=n$. Then, $\prod_{i}^{n}x_{\psi(v)} = \prod_{1}^{k-1}x_{\psi(v)}.x_{\psi(k)}.\prod_{1}^{n-k}x_{\psi(k+v)}$ $= \prod_{1}^{k-1}x_{\psi(v)}. \psi_{1}^{n-k}x_{\psi(k+v)}.x_{\psi(k)}$

Define a map $\phi$ of $(1,2, \ldots, n-1)$ into itself by the rule: $\phi(v)=\psi(v)$ if $v $\phi(v) = \psi(v+1)$ if $v \geq k$

Then, $\prod_{1}^{n} x_{\psi(v)} = \prod_{1}^{k-1}x_{\phi(v)}. \prod_{1}^{n-k}x_{\phi(k-1+v)} = \prod_{1}^{n-1}x_{\phi(v)}.x_{n}$

which by induction is equal to $x_{1}\ldots x_{n}$ as desired.

Some remarks: As a student, I used to think many a times that this proof is obvious. But it would be difficult to write it. I think this cute little proof is a good illustration of “how to prove obvious things in algebra.” 🙂

Regards,

Nalin Pithwa

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