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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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Slightly longer proof | Comment 2 of 3 |

Assume that the square root of 22 is rational.<o:p></o:p>

This means there exist integers a and b, with no common factor, such that: a/b = square root of 22.<o:p></o:p>

Squaring both sides: (a/b)^2 = 22  <o:p></o:p>

So, a^2/b^2 = 22

Hence, a^2 = 22b^2 = (2)(11)b^2  .......... eq (1)<o:p></o:p>

Since the RHS has a factor of 2, by definition it is even. Therefore, so is the LHS and so a^2 must be even. <o:p></o:p>

Since any odd number times itself is odd, it follows that if a^2 is even, then a is even.<o:p></o:p>

Then, there exists an integer k, s.t. a = 2k ....... eq. (2) <o:p></o:p>

Substitute eq. (2) into (1):   (2k)^2 = (2)(11)b^2<o:p></o:p>

                4k^2 =(2)(11)b^2<o:p></o:p>

Or             2k^2 = 11b^2     dividing both sides by 2<o:p></o:p>

Since the LHS is even, so is the RHS. On the RHS, 11 is odd which implies that b^2 must be even. And, as above, this means b is even.<o:p></o:p>

If a and b are both even, they have a common factor of 2. This contradicts the initial assumption above that they have no common factor. Therefore, the square root of 22 cannot be rational.<o:p></o:p>

This is easily generalized to any non-square integer.

  Posted by JayDeeKay on 2017-10-18 13:19:38
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