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By George Polya

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For n = 1, the identity just says 2 1=1 which is trivial. Now assume that the identity holds for an arbitrary number n. ) That is, assume that 2 1 + 3 + 5 + . . + (2n - 1) = n . The next odd number after (2n - 1) is (2n + 1). Adding this number to both sides of the identity above gives 1 + 3 + 5 + . . + (2n - 1) + (2n + 1) = n2 + (2n + 1). 2 By elementary algebra, the expression to the right of the equals sign reduces to (n + 1) . Hence you can rewrite the last identity as 2 1 + 3 + 5 + . . + (2n - 1) + (2n + 1) = (n + 1) .

I did not give any examples of multiplication above, since we never multiply hours or times of the day. But modular multiplication makes perfect sense from a mathematical point of view. As with addition, you just perform the multiplication in the usual way, but then discard all multiples of the modulus n. So, for example, with modulus 7: 2 × 3 º 6, 3 × 5 º 1. Gauss's notion of congruence is often used in mathematics, sometimes with several different moduli at the same time. When this is the case, in order to keep track of the modulus being used on each occasion, mathematicians generally write congruences like this: a º b (mod n) where n is the modulus concerned for this particular congruence.

Pn divides P; if you try to carry out such a division, you will end up with a remainder of 1—that '1' that was added on to give P in the first place. So, if P is not prime, it must be evenly divisible by some prime different from (and hence bigger than) all of p1, . . , pn. In particular, there must be a prime bigger than all of p1, . . , pn, so again, the sequence can be continued. It is interesting to observe that, when you look at the number P = p1 × p2 × . . × pn + 1 used in Euclid's proof, you don't actually know whether P is itself prime or not.

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Mathematics and Plausible Reasoning, both volumes combined by George Polya

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