# Prove that sqrt(5) is irrational?

• November 13th 2010, 09:26 PM
Thetheorycase
Prove that sqrt(5) is irrational?
My work

Suppose sqrt(5) is rational

5 = (m/n)^2

5 = m^2/n^2 or m^2 = 5n^2

Since m^2 is a multiple of 5, m must also be a multiple of 5? - proof?

m = 5k for some int k

5n^2 = (5k)^2

5n^2 = 25k^2

n = 5k^2

n and m are both multiples of 5 this contradicts the assumption sqrt is irrational

end of work

My professor didnt like what i wrote in bold. I suppose she wanted me to write a proof out for this? how would I proceed? Is this broken up into cases?

thank you
ps. Im sorry I have so many topics in this area. Im trying my hardest to get by :(
• November 13th 2010, 09:49 PM
melese
Quote:

Originally Posted by Thetheorycase
Since m^2 is a multiple of 5, m must also be a multiple of 5? - proof?

There's a result called Euclid's Lemma: If $a$ divideds $bc$ and $gcd(a,b)=1$, then $a$ divides $c$.
In particular, if $p|ts$, where $p$ is prime, then $p|t$ or $p|s$.
• November 13th 2010, 09:57 PM
Thetheorycase
Do you think i should have omitted that line altogtehr then because Euclid lemma comes in the section after?? I thought maybe she wanted me to use the quotient remainder theorem?
• November 13th 2010, 10:50 PM
Jhevon
Quote:

Originally Posted by Thetheorycase
Do you think i should have omitted that line altogtehr then because Euclid lemma comes in the section after?? I thought maybe she wanted me to use the quotient remainder theorem?

you could also prove it by the fundamental theorem of arithmetic (proof by contradiction). If you assume m is not a multiple of 5, it contradicts the uniqueness of prime factorization that the theorem asserts.

you could also do this proof another way, using the rational roots theorem. have you covered that?
• November 14th 2010, 09:16 PM
Thetheorycase
"If you assume m is not a multiple of 5, it contradicts the uniqueness of prime factorization that the theorem asserts."

This.. Could you show me how it is done?

would it be broken up into 4 cases?

1) m = 5k +1
2) m = 5k + 2
3) m = 5k + 3
4) m = 5k + 4
• November 18th 2010, 10:13 PM
Jhevon
Quote:

Originally Posted by Thetheorycase
"If you assume m is not a multiple of 5, it contradicts the uniqueness of prime factorization that the theorem asserts."

This.. Could you show me how it is done?

would it be broken up into 4 cases?

1) m = 5k +1
2) m = 5k + 2
3) m = 5k + 3
4) m = 5k + 4

well, no. not if i want to use the theorem i asserted.

We know that each integer is a product of primes, each to some positive integer power, and this product is unique for each integer. we get this from the said theorem. so now, if we take $\displaystyle m = p_1^{n_1} \cdot p_2^{n_2} \cdots p_k^{n_k}$, where the $p_i$'s are some primes, and the $n_j$'s are some powers, we would see that:

$\displaystyle m^2 = 5n^2 \implies p_1^{2n_1} \cdot p_2^{2n_2} \cdots p_k^{2n_k} = 5n^2$

Now, if we assume to the contrary that $\displaystyle m$ is not a multiple of 5, we would get that in the above equation, there is no factor of 5 on the left hand side, but there is a factor of 5 on the right hand side. this contradicts the fact that each integer has a unique prime factorization, as there is a prime factor on one side that's not on the other (that is, we can represent m with two different prime factorizations. absurd!) this contradicts the uniqueness of prime factorization theorem. hence, m must be a multiple of 5.

now of course, you can say all that much more concisely.