Mark O. answered • 12/08/17

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Hi Ashley,

First, for my proof, we will need to use the fact that √6 is irrational.

Assume that √6 is rational. This means that I can write it as √6 = a/b where a and b are in lowest terms. This means that a and b cannot both be even. If they are both even, then I can divide the numerator and denominator by 2, and they would not be in lowest terms. But, here we will say that a/b is a fraction in lowest terms. So, we are assuming that √6 is rational since a rational number can be expressed this way as a/b.

Now square both sides of the equation.

a

^{2}/ b^{2}= 6Then, we can write

6b

^{2}= a^{2}.But, the left side 6b

^{2}must be even, since anything multiplied by 6 is even. Therefore a^{2}, and hence a is even.If a is even, then I can express it as some 2c, where c is an integer. So,

6b

^{2}= a^{2}becomes 6b^{2}= (2c)^{2}or 6b^{2}= 4c^{2}We can divide both sides of this last equation by 2 and get

3b

^{2}= 2c^{2}So, b

^{2}and hence b must be even since 3 is odd and the right side of the equation is definitely even since it is multiplied by 2.We have now shown that a and b are both even. But, we said that for √6 to be rational, a and b cannot both be even. This is a contradiction. So, √6 is irrational.

Now, let's get to your problem.

Let (√2 + √3) = r, and we claim that r is rational.

Let's square both sides of this equation.

We get 5 + 2√6 = r

^{2}or

r

^{2}- 5 = 2√6or

(r

^{2}- 5) / 2 = √6We assumed r is rational, so r

^{2}is rational. 5 is certain rational, since 5 = 5/1. In the same way 2 is rational. We can divide rational r^{2}- 5 by rational 2 and we should get a rational. So, the right side of this last equation should be rational since the quotient of the left is definitely rational. But, we already proved above that √6 is irrational, which is a contradiction. Therefore, r = √2 + √3 is irrational.
Ashley H.

12/08/17