- #1

njkid

- 22

- 0

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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- Thread starter njkid
- Start date

- #1

njkid

- 22

- 0

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.

- #2

Atomos

- 165

- 0

and what does this have to do with geometry?

- #3

amcavoy

- 665

- 0

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**.

[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

Last edited:

- #4

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

"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 a^{2}= 2b^{2} showing that a^{2} 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.

p^{2}= (2n+1)^{2}= 4n^{2}+ 4n+ 1= 2(2n^{2}+2n)+1.

Since 2n^{2}+ 2n is an integer, p^{2} is of the form 2m+1 (m= 2n^{2}+2n) and so is odd.

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

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

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 a

Crucial point: the square of an

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

p

Since 2n

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

Last edited by a moderator:

- #5

njkid

- 22

- 0

HallsofIvy said:"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 anirrationalnumber 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 a^{2}= 2b^{2}showing that a^{2}iseven.

Crucial point: the square of anoddinteger is alwaysodd:

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

p^{2}= (2n+1)^{2}= 4n^{2}+ 4n+ 1= 2(2n^{2}+2n)+1.

Since 2n^{2}+ 2n is an integer, p^{2}is of the form 2m+1 (m= 2n^{2}+2n) and so is odd.

Do you see why knowing that tells us that a must beeven?

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

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