Prove the Square Root of 2 is irrational

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To prove that the square root of 2 is irrational, one must assume it can be expressed as a fraction a/b, where a and b are natural numbers. By squaring both sides, it follows that a² = 2b², indicating that a² is even, which implies that a must also be even. If a is even, it can be expressed as 2k for some integer k, leading to a contradiction when substituting back into the original equation. This contradiction shows that a and b cannot both be integers, thus proving that the square root of 2 is irrational. The discussion emphasizes the importance of understanding even and odd integers in this proof.
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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.
 
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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?
 
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.
 
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"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?
 
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HallsofIvy said:
"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!
 
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