Prove the Square Root of 2 is irrational

[njkid]

This is Algebra 2 question...

I have to prove that the square root of 2 is irrational...

First we must assume that

sqrt (2) = a/b

I never took geometry and i don't know proofs...

Please help me.

Thank you.

 

[Atomos]

a rational number is of form a/b where a and be are mutually prime. I will give you a hint: you must prove that and and b cannot possibly be mutually prime.

and what does this have to do with geometry?

 

[amcavoy]

You're off to a good start. Let "a" and "b" be natural numbers.

$$\sqrt{2}=\frac{a}{b}\implies b\sqrt{2}=a$$

Now, how could "a" be a natural number? Well, "b" would be some multiple of $\sqrt{2}$. This in turn, would mean that "b" isn't a natural number.

Can you see where this is going? You need to prove this by contradiction.

 

[HallsofIvy]

"Now, how could "a" be a natural number? Well, "b" would be some multiple of $\sqrt{2}$. This in turn, would mean that "b" isn't a natural number."
How does that follow? Saying b*(1/2) , for example, equals a natural number does not imply that b isn't a natural number! Of course 1/2 isn't an irrational number but the whole point here is to prove that $\sqrt{2}$ is irrational.

Better to note that if $\frac{a}{b}= \sqrt{2}$ then, squaring both sides, $\frac{a^2}{b^2}= 2$ so that a²= 2b² showing that a² is even.

Crucial point: the square of an odd integer is always odd:

If p is an odd integer, then it can be written 2n+ 1 where n is any integer.

p²= (2n+1)²= 4n²+ 4n+ 1= 2(2n²+2n)+1.

Since 2n²+ 2n is an integer, p² is of the form 2m+1 (m= 2n²+2n) and so is odd.

Do you see why knowing that tells us that a must be even?

 

[njkid]

"Now, how could "a" be a natural number? Well, "b" would be some multiple of $\sqrt{2}$. This in turn, would mean that "b" isn't a natural number."
How does that follow? Saying b*(1/2) , for example, equals a natural number does not imply that b isn't a natural number! Of course 1/2 isn't an irrational number but the whole point here is to prove that $\sqrt{2}$ is irrational.

Better to note that if $\frac{a}{b}= \sqrt{2}$ then, squaring both sides, $\frac{a^2}{b^2}= 2$ so that a²= 2b² showing that a² is even.

Crucial point: the square of an odd integer is always odd:

If p is an odd integer, then it can be written 2n+ 1 where n is any integer.

p²= (2n+1)²= 4n²+ 4n+ 1= 2(2n²+2n)+1.

Since 2n²+ 2n is an integer, p² is of the form 2m+1 (m= 2n²+2n) and so is odd.

Do you see why knowing that tells us that a must be even?

Wow! You are so good at teaching! Thank you everybody!