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√22 (Posted on 2017-10-18) Difficulty: 4 of 5
It easy to show that √2 is an irrational number

So is the square root of any non-square number.

Prove that √22 is an irrational number, providing a reasoning applicable to any random-chosen non-square integer.

See The Solution Submitted by Ady TZIDON    
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Solution short general proof Comment 3 of 3 |
Let n be an integer that's not a perfect square. At least one of its prime factors must appear within it as a factor an odd number of times in order for n not to be a perfect square.

Suppose sqrt(n) is rational so that sqrt(n) = a/b where a and b are whole numbers and a/b is in lowest terms, that is, GCD(a,b)=1. Then n = a^2/b^2 or a^2 = n * b^2.

The LHS is a perfect square, so every prime factor of the LHS appears as an even power.  Every prime in b^2 similarly appears as an even power; but n contains at least one prime that's not to an even power. Thus the combined n * b^2 has a prime to an odd power. This is a contradiction to our assumptions, which were only that sqrt(n) is rational and n is not a perfect square.

  Posted by Charlie on 2017-10-18 15:28:53
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