updated: 2022-01-23_12:32:30-05:00
sqrt(2) is irrational
A number is rational if it is a fraction of form $m/n$, where m and n are integers
Proof by Contradiction:
Assume $\sqrt(2)$ is rational
sqrt(2) = m/n where m and n are integers
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m and n are divisible by some integer > 1
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reduce fraction:
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if reduced as much as possible, at least one is odd
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sqrt(2)=m/n
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square both sides
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2 = m^2/n^2
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multiply n^2 on both sides
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2n^2 = m^2
- implies m^2 is even
- if m^2 is even, m is also even
- (square of an odd number is always odd)
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Since m is even we can write it as m = 2k
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substitute m into 2n^2 = m^2
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2n^2 = (2k)^2
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n^2 = 2k^2
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therefore n is even
both m and n are even
which contradicts one having to be odd