Irrationality of Difference of two numbers

[smiles988]

Proof of Irrationality

How can I prove that the square root of 8 minus the square root of 3 is an irrational number using the fact that the square root of 3 is an irrational number? I know I need to use a proof by contradiction, but I am stuck after that.

 

[smiles988]

Homework Statement

\sqrt{8}-\sqrt{3} is an irrational number.
Use the fact that \sqrt{3} is an irrational number to prove the following theorem.

Homework Equations

A rational number can be written in the form \frac{p}{q} where p and q are integers in lowest terms.

The Attempt at a Solution

I know that I need to use a proof by contradiction to solve this problem. Therefore, we would assume that \sqrt{8}-\sqrt{3} is an rational number and that \sqrt{3} is an irrational number and try and reason to a contradiction. I am stuck and don't know how to get to the contradiction.

 

[MathematicalPhysicist]

well it's actually asking to prove that
(sqrt(2)+sqrt(3))+sqrt(2)
is irrational.
assume it equals p/q where (p,q)=1
then multiply by sqrt(8)-sqrt(3)
youll get that sqrt(8)-sqrt(3)=5q/p
and you also have sqrt(8)+sqrt(3)=p/q
so you get that 4sqrt(2)=5q/p+p/q
which yields: sqrt(2)=(5q/p+p/q)/4 which is a contradiction.
you could have easily have done it with sqrt(3) instead but it doesn't matter.

 

[MathematicalPhysicist]

well, i had mistaken sqrt(8)-sqrt(3) with sqrt(8)+sqrt(3) but it doenst matter the same idea will work as well as in this case.

 

[matt grime]

Suppose that it is, then this implies that the square root of 8 is what?

 

[WWGD]

Just to add something to review my number theory.Another idea:

I think a nice general result is that for x a pos. integer, sqr(x) is rational iff x is a perfect square (an integer, of course). Think x=a^2/b^2 , so a^2x=b^2 . Then think of what the factorization of x needs to satisfy in order for a^2x to be a perfect square.

x=p_1^e_1...p_ne^n .



Then , re your problem, think of what happens when you square your expression.

 

[HallsofIvy]

I believe this is the third time you have posted this same question in a separate thread!