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Prove the Square Root of 2 is irrational

by njkid
Tags: irrational, prove, root, square
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njkid
#1
Sep8-05, 09:11 PM
P: 22
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 dont know proofs...

Please help me.

Thank you.
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Atomos
#2
Sep8-05, 09:33 PM
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P: 159
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
#3
Sep8-05, 10:12 PM
P: 666
You're off to a good start. Let "a" and "b" be natural numbers.

[tex]\sqrt{2}=\frac{a}{b}\implies b\sqrt{2}=a[/tex]

Now, how could "a" be a natural number? Well, "b" would be some multiple of [itex]\sqrt{2}[/itex]. 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
#4
Sep9-05, 06:02 AM
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Prove the Square Root of 2 is irrational

"Now, how could "a" be a natural number? Well, "b" would be some multiple of [itex]\sqrt{2}[/itex]. 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 [itex]\sqrt{2}[/itex] is irrational.

Better to note that if [itex]\frac{a}{b}= \sqrt{2}[/itex] then, squaring both sides, [itex]\frac{a^2}{b^2}= 2[/itex] 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
#5
Sep9-05, 01:50 PM
P: 22
Quote Quote by HallsofIvy
"Now, how could "a" be a natural number? Well, "b" would be some multiple of [itex]\sqrt{2}[/itex]. 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 [itex]\sqrt{2}[/itex] is irrational.

Better to note that if [itex]\frac{a}{b}= \sqrt{2}[/itex] then, squaring both sides, [itex]\frac{a^2}{b^2}= 2[/itex] 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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