Algebra is symbolic manipulation though painstaking or conscientious :-)

Of course, I have oversimplified the meaning of algebra. ๐Ÿ™‚

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<k

\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.” ๐Ÿ™‚

Regards,

Nalin Pithwa

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