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Proof that the square root of any non-square number is irrational.

First let's look at the proof that the square root of 2 is irrational. First, let's suppose that the square root of two is rational. Therefore, it can be expressed as a fraction: \frac{a}{b} . Then let's suppose that \frac{a}{b} is in lowest terms, meaning a,b are relative primes, meaning their greatest common factor is 1.

So far, \sqrt2=\frac{a}{b} . Let's square both sides. 2=\frac{a^2}{b^2} . Then multiply both sides by b^2 , 2b^2=a^2 meaning, and you can test this out, a^2 has to be divisible by two, or in other words, it is even.

Let's keep on going. I'm going to replace a with 2r , because since a is even, it is the product of 2 and some number (r) 2b^2=(2r)^2\ ;\ 2b^2=4r^2 . Divide both sides by 2: b^2=2r^2 , and we are left with a situation where b has to be divisible by 2. Meaning both a,b are even.

This means that the square root of 2 is irrational because you can replace a with \frac{a}{2} and b with \frac{b}{2} and get the same results.

Let's try it with a square number now, 4. \sqrt4=\frac{a}{b}\ ;\ 4=\frac{a^2}{b^2}\ ;\ 4b^2=a^2 now notice that that this doesn't mean that a is divisible by 4. It means it is even, because the situation could be that a is 2. While with 2 and any other non-square number, there is no possible whole number a that when squared equals that number. Examples: 35b^2=a^2 . a has to be divisible by 35. Let's try 81, a square number. 81b^2=a^2 . a doesn't have to be divisible by 81, but only by 9 because a could be 9 and 9^2=81 .

Let's continue. Replacing a with 2r . 4b^2=(2r)^2\ ;\ 4b^2=4r^2 . b^2=r^2 We have no proof that b is divisible by 2 so we can assume(even though we already know) that the square root of 4 is rational. Now for any positive intiger, you will end up with b^2=nr^2 where n is the square root of whatever number you are trying to prove the square root of is rational.

In conclusion, that is the proof that the square root of any positive integer that isn't a square number is irrational

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