proof:
Assume sqrt(3)= a/b where a,b are integers and b cannot equal 0. (sqrt(3)=(a/b)^2 turns 3=a^2/b^2 and 3b^2=a^2. We know that 3|a^2 iff 3|a. Using this theorem we know that 3(b^2)=a^2 must be a multiple of 3|a. 3|a then becomes 3r=a where r is some integer. Then squaring both side 3r=a becomes 9r^2=a^2. Since 3b^2=a^2 that means that 3b^2=9r^2 which simplifies into b/r=sqrt(3).
Since b/r is a rational number this is a contradiction
I think its right not sure though.
I havent learned the floor function yet. The book tells me to use this as a lemma: 3|a^2 if and only if 3|a. Since 3(b^2)=a^2 and 9r^2=a^2 from squaring 3|a I can just set them equal to each other and it becomes
3b^2=9r^2? and then show that b/r=sqrt(3) and b/r is rational which is a contradiction.
Hello, bonfire09!
I'll give you the standard proof.
Assume that is rational.
We have:
Square the equation: . .[4]
We see that is a multiple of 3; hence is a multiple of 3.
Let
Then [4] becomes: .
We see that is a multiple of 3; hence is a multiple of 3.
But if is a multiple of 3 and is a multiple of 3,
. . then the fraction is not reduced to lowest terms.
We have arrived at a contradiction.
Therefore, is not rational.
Here is my new revised proof then
Assume sqrt(3)= a/b where a,b are integers and b cannot equal 0. We can also assume that a/b has been reduced to lowest terms. Sqrt(3)=(a/b)^2 turns 3=a^2/b^2 and 3b^2=a^2. We know that 3|a^2 iff 3|a. Using this theorem we know that 3(b^2)=a^2 must be a multiple of 3|a. 3|a then becomes 3r=a where r is some integer. Then squaring both side 3r=a becomes 9r^2=a^2. Since 3b^2=a^2 that means that 3b^2=9r^2 which implies 3r^2=b^2 where r^2 is an integer. By the previous thereom stated we know that if 3|b^2 then 3|b. Since a and b both share a divisor of 3 that means it has not been reduced to its lowest terms. Hence by contradiction sqrt(3) is irrational.