Prove: 3(v2)-5 is irrational given that v2 is irrational? Please Help!?

Question

I have tried direct proof and proof by contradiction.

Prove: 3(√2)-5 is irrational given that √2 is irrational

Thats pretty much it, I have had a few starts but they dont go anywhere.  

Answer 1

Mind you I'm not an expert on mathematical proofs but here's a simple take on it:

 

Suppose that 3*(√2)-5 is a rational number, it can therefore be expressed as a ratio of integers m and n:

 

3(√2)-5=m/n

 

Adding 5 to both sides:

3(√2)=m/n+5=m/n+5n/n=(m+5n)/n

 

Dividing both sides by 3:

√2=(m+5n)/(3n)

 

Now since both m and n are integers, then both terms (m+5n) and 3n are also integers which means √2 can then be expressed as a ratio of integers (ie it's a rational number). Since we are explicitly told that √2 is not a rational number and therefore cannot be expressed as a ratio of integers, this is obviously false. So therefore, the initial assumption that 3(√2)-5 is a rational number is also false so it must be an irrational number.