Is the Square Root of 2 Rational?

[annoymage]

sorry, I am using phone, owho i hope you can get it.

suppose its rational. in the form a/b gcd(a,b)=1
and so and so and so and then they concluded that a^2=2 is a contradiction.

but i cannot see what it conradict the assumption. help

 

[Dick]

a^2=2 isn't the contradiction. I'm guessing you didn't read the whole proof. Try that and then check back if you still don't get it.

 

[annoymage]

i'll denote a^2 is a2

then a2=2b2
hence b l a2
if b>1 <skipped> we have contradiction. so b=1. hence a2=2, a contradiction. so squareroot 2 is not rational.

we can't conclude, a2=2, a contradiction right? i can't concentrate on other lecture because this is bothering me. and of course i know how to prove it in other way

 

[Dick]

i'll denote a^2 is a2

then a2=2b2
hence b l a2
if b>1 <skipped> we have contradiction. so b=1. hence a2=2, a contradiction. so squareroot 2 is not rational.

we can't conclude, a2=2, a contradiction right? i can't concentrate on other lecture because this is bothering me. and of course i know how to prove it in other way

Ok, so b | a^2. Now why is b>1 supposed to be a contradiction? Why did you write <skipped>?

 

[annoymage]

sorry i use phone, now I'm on computer,

so if b>1, then there exist a prime number p such that p l b, so p l a^2, then p l a, hence gcd(a,b)=p>1, a contradiction, so b must be 1. so that prove is inconclusive right?

btw, if you have time, can you response in this https://www.physicsforums.com/showthread.php?t=424018

thanks ^^

 

[Dick]

sorry i use phone, now I'm on computer,

so if b>1, then there exist a prime number p such that p l b, so p l a^2, then p l a, hence gcd(a,b)=p>1, a contradiction, so b must be 1. so that prove is inconclusive right?

btw, if you have time, can you response in this https://www.physicsforums.com/showthread.php?t=424018

thanks ^^

That looks ok to me. Except you should write gcd(a,b)>=p if all you know is that p | a and p | b.

 

[annoymage]

so do you mean a^2=2 is a contradiction?

 

[Dick]

so do you mean a^2=2 is a contradiction?

Well, yes. Contrary to what I said in post 2. I wasn't following your proof. a^2=2 is a contradiction because a is an integer and there is no integer whose square is 2. You are basically proving that the only integers who have rational square roots are the perfect squares.

 

[annoymage]

there is no integer whose square is 2.

thankssss