Help me with proof by contradiction pls

Question

I need to prove the fact that 2^(1/2)+3^(1/3) is irrational. Can someone help with this pls

Answer 1

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.

 

a² / b² = 6

 

Then, we can write

 

6b² = a².

 

But, the left side 6b² must be even, since anything multiplied by 6 is even. Therefore a², and hence a is even.

 

If a is even, then I can express it as some 2c, where c is an integer. So,

6b² = a² becomes 6b² = (2c)² or 6b² = 4c²

 

We can divide both sides of this last equation by 2 and get

 

3b² = 2c²

 

So, b² 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 = r²

 

or

 

r² - 5 = 2√6

 

or

 

(r² - 5) / 2 = √6

 

We assumed r is rational, so r² is rational. 5 is certain rational, since 5 = 5/1. In the same way 2 is rational. We can divide rational r² - 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.