proof of an irrational number

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Guest
hey can somebody help me with proving that the square root of 3 is irrational?
This is how far I got:

(sqrt 3 b)^3 = (a)^3
3b^3=a^3
now what?
 
Do you know how to show that sqrt[2] is irrational? If not, study that proof for advice. If so, then model this proof after that one. The only difference will be in checking the square. Instead of "either it's a multiple of 2, or one more than a multiple of 2", you'll have to check "either it's a multiple of 3, or one more, or two more". Etc.

Eliz.
 
ok so i looked at the proof and i dont understand this part:

how did
2b^2=a^2
2b^2=4a^2?

would it be like this:

3b^3=6a^3
or
3b^3=9a^3
and then you would end up with 3/6 or 3/9which is not in its lowest form and therefore irrtational?
 
Brit412 said:
how did
2b^2=a^2
2b^2=4a^2?
Since I can't see the particular proof you're looking at, I can't say definitely how the author got to this point. But there should have been some discussion of what a and b were, and the evenness or oddness of at least one of them.

Eliz.
 
Hello, Brit412

hey can somebody help me with proving that the square root of 3 is irrational?

This is how far I got:
(sqrt 3 b)^3 = (a)^3 . . . . Why are you cubing?
3b^3=a^3 . . . . This is not true!
now what?

The standard proof goes like this . . .

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

Then: ab=3\displaystyle \,\frac{a}{b}\:=\:\sqrt{3}\, where a\displaystyle a and b\displaystyle b are positive, relative prime integers.
. . That is, the fraction is already reduced to lowest terms.

And we have: .a=b3\displaystyle a\:=\:b\sqrt{3}

Square both sides: .a2=3b2\displaystyle a^2\:=\:3b^2 .[1]


The right side is a multiple of 3, hence a2\displaystyle a^2 is a multiple of 3.
If a2\displaystyle a^2 is a multiple of 3, then a\displaystyle a is a multiple of 3.
. . Hence: .a=3h\displaystyle a \,=\,3h for some integer h\displaystyle h

Substitute into [1]: .(3h)2=3b2        9h2=3b2        3h2=b2\displaystyle (3h)^2\,=\,3b^2\;\;\Rightarrow\;\; 9h^2\,=\,3b^2\;\;\Rightarrow\;\; 3h^2\,=\,b^2

The left side is a multiple of 3, hence b2\displaystyle b^2 is a multiple of 3.
If b2\displaystyle b^2 is a multiple of 3, then b\displaystyle b is a multiple of 3.
. . Hence: .b=3k\displaystyle b\,=\,3k for some integer k\displaystyle k

But if a=3h\displaystyle a\,=\,3h and b=3k\displaystyle b\,=\,3k, the fraction ab=3h3k\displaystyle \frac{a}{b}\,=\,\frac{3h}{3k} is not reduced to lowest terms.


We have reached a contradiction. .Hence, our assumption was incorrect.

Therefore: 3\displaystyle \sqrt{3} is irrational.

 
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