# Thread: prove sqrt(3) is irrational

1. ## prove sqrt(3) is irrational

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.

2. ## Re: prove sqrt(3) is irrational

Originally Posted by bonfire09
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 will comment on your proof because I do not follow it.
There is a standard proof.
If the integer $n$ is not a square then $\sqrt n$ is irrational.
Prove that by using the floor function.

3. ## Re: prove sqrt(3) is irrational

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.

4. ## Re: prove sqrt(3) is irrational

Originally Posted by bonfire09
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.
You have an out-of-date textbook.

5. ## Re: prove sqrt(3) is irrational

Hello, bonfire09!

I'll give you the standard proof.

$\text{Prove that }\sqrt{3}\text{ is irrational.}$

Assume that $\sqrt{3}$ is rational.

We have: $\sqrt{3} \,=\,\frac{a}{b}\;\text{ where }\:\begin{Bmatrix}(1) & a\text{ and }b\text{ are integers.} \\ (2) & \text{The denominator }b\text{ is not zero.} \\ (3) & \text{The fraction }\dfrac{a}{b}\text{ is reduced to lowest terms.}\end{Bmatrix}$

Square the equation: . $3 \:=\:\frac{a^2}{b^2} \quad\Rightarrow\quad 3b^2 \:=\:a^2$ .[4]

We see that $a^2$ is a multiple of 3; hence $a$ is a multiple of 3.

Let $a = 3r$

Then [4] becomes: . $3b^2 \:=\:9r^2\quad\Rightarrow\quad b^2 \:=\:3r^2$

We see that $b^2$ is a multiple of 3; hence $b$ is a multiple of 3.

But if $a$ is a multiple of 3 and $b$ is a multiple of 3,
. . then the fraction $\frac{a}{b}$ is not reduced to lowest terms.

We have arrived at a contradiction.
Therefore, $\sqrt{3}$ is not rational.

6. ## Re: prove sqrt(3) is irrational

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.

7. ## Re: prove sqrt(3) is irrational

Looks correct except for the sentence "Using this theorem we know that 3(b^2)=a^2 must be a multiple of 3|a"
Maybe modify it to be something like "From 3(b^2)=a^2, we know that 3|a^2, and so this theorem shows that 3|a."