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f24u7
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I saw a proof on a textbook, but I don't understand why m and n should not both be even?
Prove p^2 =2 then p is irrational
(1) p^2 = 2
is not satisfied by any rational p. If there were such a p, we could write p = m/n
where m and n are integers that are not both even.
Then (1) implies
(2) m^2=2n^2
This shows that m^2 is even.
Hence m is even (if m were odd, m 2 would be odd),and so m^2 is divisible by 4.
It follows that the right side of (2) is divisible by 4,
so that n is even, which implies that n is even.
The assumption that (1) holds thus leads to the conclusion that both m
and n are even, contrary to our choice of m and n. Hence (1) is impossible for
rational p.
Thanks in advance
Prove p^2 =2 then p is irrational
(1) p^2 = 2
is not satisfied by any rational p. If there were such a p, we could write p = m/n
where m and n are integers that are not both even.
Then (1) implies
(2) m^2=2n^2
This shows that m^2 is even.
Hence m is even (if m were odd, m 2 would be odd),and so m^2 is divisible by 4.
It follows that the right side of (2) is divisible by 4,
so that n is even, which implies that n is even.
The assumption that (1) holds thus leads to the conclusion that both m
and n are even, contrary to our choice of m and n. Hence (1) is impossible for
rational p.
Thanks in advance
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