G
Guest
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?
This is how far I got:
(sqrt 3 b)^3 = (a)^3
3b^3=a^3
now what?
- Joined
- Feb 4, 2004
- Messages
- 16,582
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.
Eliz.
G
Guest
Guest
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?
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?
- Joined
- Feb 4, 2004
- Messages
- 16,582
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.Brit412 said:
Eliz.
Hello, Brit412
The standard proof goes like this . . .
Assume that \(\displaystyle \sqrt{3}\) is rational.
Then: \(\displaystyle \,\frac{a}{b}\:=\:\sqrt{3}\,\) where \(\displaystyle a\) and \(\displaystyle b\) are positive, relative prime integers.
. . That is, the fraction is already reduced to lowest terms.
And we have: .\(\displaystyle a\:=\:b\sqrt{3}\)
Square both sides: .\(\displaystyle a^2\:=\:3b^2\) .[1]
The right side is a multiple of 3, hence \(\displaystyle a^2\) is a multiple of 3.
If \(\displaystyle a^2\) is a multiple of 3, then \(\displaystyle a\) is a multiple of 3.
. . Hence: .\(\displaystyle a \,=\,3h\) for some integer \(\displaystyle h\)
Substitute into [1]: .\(\displaystyle (3h)^2\,=\,3b^2\;\;\Rightarrow\;\; 9h^2\,=\,3b^2\;\;\Rightarrow\;\; 3h^2\,=\,b^2\)
The left side is a multiple of 3, hence \(\displaystyle b^2\) is a multiple of 3.
If \(\displaystyle b^2\) is a multiple of 3, then \(\displaystyle b\) is a multiple of 3.
. . Hence: .\(\displaystyle b\,=\,3k\) for some integer \(\displaystyle k\)
But if \(\displaystyle a\,=\,3h\) and \(\displaystyle b\,=\,3k\), the fraction \(\displaystyle \frac{a}{b}\,=\,\frac{3h}{3k}\) is not reduced to lowest terms.
We have reached a contradiction. .Hence, our assumption was incorrect.
Therefore: \(\displaystyle \sqrt{3}\) is irrational.
The standard proof goes like this . . .
Assume that \(\displaystyle \sqrt{3}\) is rational.
Then: \(\displaystyle \,\frac{a}{b}\:=\:\sqrt{3}\,\) where \(\displaystyle a\) and \(\displaystyle b\) are positive, relative prime integers.
. . That is, the fraction is already reduced to lowest terms.
And we have: .\(\displaystyle a\:=\:b\sqrt{3}\)
Square both sides: .\(\displaystyle a^2\:=\:3b^2\) .[1]
The right side is a multiple of 3, hence \(\displaystyle a^2\) is a multiple of 3.
If \(\displaystyle a^2\) is a multiple of 3, then \(\displaystyle a\) is a multiple of 3.
. . Hence: .\(\displaystyle a \,=\,3h\) for some integer \(\displaystyle h\)
Substitute into [1]: .\(\displaystyle (3h)^2\,=\,3b^2\;\;\Rightarrow\;\; 9h^2\,=\,3b^2\;\;\Rightarrow\;\; 3h^2\,=\,b^2\)
The left side is a multiple of 3, hence \(\displaystyle b^2\) is a multiple of 3.
If \(\displaystyle b^2\) is a multiple of 3, then \(\displaystyle b\) is a multiple of 3.
. . Hence: .\(\displaystyle b\,=\,3k\) for some integer \(\displaystyle k\)
But if \(\displaystyle a\,=\,3h\) and \(\displaystyle b\,=\,3k\), the fraction \(\displaystyle \frac{a}{b}\,=\,\frac{3h}{3k}\) is not reduced to lowest terms.
We have reached a contradiction. .Hence, our assumption was incorrect.
Therefore: \(\displaystyle \sqrt{3}\) is irrational.