Proving something is irrational

[ver_mathstats]

Homework Statement

Prove that for all x ∈ ℝ, at least one of √3 - x and √3 +x is irrational.

Homework Equations

The Attempt at a Solution

I understand how to prove that √3 is a irrational number by proof by contradiction. However I am not sure how to prove this one.

Would I have to equate √3 - x = a/b and √3 + x = a/b and prove by contradiction?

Thank you.

 

[PeroK]

Homework Statement

Prove that for all x ∈ ℝ, at least one of √3 - x and √3 +x is irrational.

Homework Equations

The Attempt at a Solution

I understand how to prove that √3 is a irrational number by proof by contradiction. However I am not sure how to prove this one.

Would I have to equate √3 - x = a/b and √3 + x = a/b and prove by contradiction?

Thank you.

That would be a good start. Although the two numbers can't both be equal to the same $a/b$.

 

[fresh_42]

If you see something with $a-b$ and $a+b$ it is always an idea to multiply them. Try a proof by contradiction, i.e. assume both were rational.

 

[PeroK]

If you see something with $a-b$ and $a+b$ it is always an idea to multiply them. Try a proof by contradiction, i.e. assume both were rational.

Or add them!

 

[ver_mathstats]

If you see something with $a-b$ and $a+b$ it is always an idea to multiply them. Try a proof by contradiction, i.e. assume both were rational.

So I'd multiply (√3-x)(√3+x) and then equate it to a/b?

 

[fresh_42]

So I'd multiply (√3-x)(√3+x) and then equate it to a/b?

@PeroK's hint is the better idea: what do you get if you add / subtract the two?

 

[ver_mathstats]

@PeroK's hint is the better idea: what do you get if you add / subtract the two?

Okay so when I add then I obtain √3 + √3 or 2√3 and then this is what we equate to a/b? Thank you.

 

[ver_mathstats]

Or add them!

Thank you. I'd get 2√3 = a/b?

 

[fresh_42]

The proof starts with - as you already suggested: Assume $x-\sqrt{3}$ and $x+\sqrt{3}$ were both rational. Then their sum and difference ...