## Square roots and such

For the discussion of math. Duh.

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mcmesher
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Joined: Mon Dec 13, 2010 2:43 am UTC

### Square roots and such

Is there any proof that the square (or cube etc) roots of numbers that don't have integer square (or cube etc) roots are irrational?
Math is the coolest sport.

mdyrud
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### Re: Square roots and such

Nope, because that isn't true. sqrt(9/4)=3/2.
Aside from that type of case, I think I have seen a proof, but I can't think of it now. If the numerator and denominator are not perfect squares, it will be irrational.

mcmesher
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### Re: Square roots and such

Thanks for clarifying. I think I meant "square (or cube etc) roots of integers", but it would extend to fractions with perfect square/cube/etc numerators ad denominators. Positive, of course.
Math is the coolest sport.

jestingrabbit
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### Re: Square roots and such

It is true that "if n is an integer and not a perfect square, then [imath]\sqrt{n}[/imath] is irrational." To prove it, have a good look at the proof for [imath]\sqrt{2}[/imath] and carefully generlise it.
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mcmesher
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### Re: Square roots and such

As it turns out, I am not familiar with such a proof. I am only a freshman in high school, so I not had much opportunity to encounter higher math.
Math is the coolest sport.

jestingrabbit
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### Re: Square roots and such

Fair enough. Any of the first three proofs here should be pretty easily generalisable.

However, you might want to try writing a proof that the square root of 2 is irrational by yourself first, then see how yours looks compared to the other proofs, just to try getting a start with proving things, a skill that will come in handy if you pursue further study in maths later.
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mdyrud
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### Re: Square roots and such

Definitely try and prove it yourself. It is a pretty nifty proof.
Hint:
Spoiler:
Go for proof by contradiction. Assume that the square root of two is rational. See if stuff explodes.

WhiteRAZOR
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### Re: Square roots and such

It's something you might encounter in first year mathematics in university. You're thinking ahead!

I found it pretty cool... especially at the end where it's like "that doesn't work... all the math is right... that must mean we messed up the start! QED".

Yakk
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### Re: Square roots and such

So, a useful results for proving this:
Unique Factorization: Let n be a positive integer. Then n = product pini (possibly an empty product, which by definition equals 1), where each pi is a unique prime and ni is a positive integer. Furthermore, this entire product is unique up to reordering.

Basically, this says that any positive integer can be expressed as a unique product of prime numbers.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

polymer
Posts: 125
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### Re: Square roots and such

Another fun thm is the rational roots test. It isn't difficult to prove the root of primes are irrational using it.

Spoiler:
Thm: Consider the polynomial [imath]a_nx^n+a_{n-1}x^{n-1} + \ldots+ a_1x + a_{0} = 0[/imath] with [imath]a_{n} \neq 0[/imath] and [imath]a_i \in \mathbf{Z}[/imath] for all i. If there exists a rational root [imath]r = \frac{p}{q} \ p,q \in \mathbf{Z}[/imath] with no common factors, then [imath]p[/imath] divides [imath]a_0[/imath] and [imath]q[/imath] divides [imath]a_n[/imath].

Proof: p divides [imath]a_0[/imath].
$\begin{eqnarray} 0 &=& a_nx^n+a_{n-1}x^{n-1} + \ldots + a_1x+a_{0}\\ 0 &=&a_n\left(\frac{p}{q}\right)^n+a_{n-1}\left(\frac{p}{q}\right)^{n-1} + \ldots + a_{1}\left(\frac{p}{q}\right) + a_{0} \\ 0 &=& a_np^n+a_{n-1}qp^{n-1} + \ldots + a_{1}q^{n-1}p + a_{0}q^n \\ -a_{0}q^n &=& p\left(a_np^{n-1}+a_{n-1}qp^{n-2} + \ldots + a_{1}q^{n-1}\right) \\ \text{p} &\text{ is }& \text{a factor of }a_0q^n \text{, and p and q have no common factors; therefore, p is a factor of }a_0\text{.}\\ &&\text{This is just the definition of divides, so p divides }a_0\text{.} \end{eqnarray}$

You might try and see if you can prove the second half of the statement. The theorem is very useful for testing whether basic polynomials like [imath]x^2 - 2 = 0[/imath] have rational solutions.
Last edited by polymer on Tue Feb 01, 2011 6:44 pm UTC, edited 11 times in total.

Qaanol
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### Re: Square roots and such

polymer wrote:Another fun thm is the rational roots test. It isn't difficult to prove primes are irrational using it.

Spoiler:
Thm: Consider the polynomial [imath]a_nx^n+a_{n-1}x^{n-1} + \ldots+ a_1x + a_{0} = 0[/imath] with [imath]a_{n} \neq 0[/imath] and [imath]a_i \in \mathbf{Z}[/imath] for all i. If there exists a rational root [imath]r = \frac{p}{q} \ p,q \in \mathbf{Z}[/imath] with no common factors, then [imath]p[/imath] divides [imath]a_0[/imath] and [imath]q[/imath] divides [imath]a_n[/imath].

Proof: p divides [imath]a_0[/imath].
$\begin{eqnarray} a_nx^n+a_{n-1}x^{n-1} + \ldots + a_1x+a_{0} &=& 0\\ a_n\left(\frac{p}{q}\right)^n+a_{n-1}\left(\frac{p}{q}\right)^{n-1} + \ldots + a_{1}\left(\frac{p}{q}\right) + a_{0} &=& 0\\ p\left(a_n\left(\frac{p^{n-1}}{q^n}\right)+a_{n-1}\left(\frac{p^{n-2}}{q^{n-1} }\right)+ \ldots + \left(\frac{a_{1}}{q}\right)\right)&=& -a_0\\ \text{p is a factor of }a_0 \text{ therefore, p divides }a_0 \end{eqnarray}$

You might try and see if you can prove the second half of the statement. The theorem is very useful for testing whether basic polynomials like [imath]x^2 - 2 = 0[/imath] have rational solutions.

You do not quite have a proof there.

Spoiler:
You have no guarantee that the thing multiplying p is an integer. You want to multiply both sides by [imath]q^n[/imath] so you have integers all around. Then it follows that p divides [imath]a_0 \cdot q^n[/imath]. But p and q are relatively prime, so p divides [imath]a_0[/imath].
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mike-l
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### Re: Square roots and such

polymer wrote:It isn't difficult to prove primes are irrational using it.

I assume you mean 'roots of primes'?
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Entropyst
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### Re: Square roots and such

You could generalize something along the lines of the square root of two is irrational, or you could turn it into a conditional statement and see what that tells us. So another way of asking the question would be to examine the conditional statement "Given that m is an integer, if m is not a perfect square, then the square root of m is irrational." From conditional statements, we know that P implies Q is logically equivalent to ~Q implies ~P. In other words, if we can show that the statement "If the square root of m is rational, then m is a perfect square" to be true, then we have proven what we want to know. This is a much easier approach in my opinion.

polymer
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### Re: Square roots and such

yea...you caught me before I corrected myself. This is why one shouldn't do math at 12:30...I'll correct it so the post isn't confusing.

Qaanol wrote:
You do not quite have a proof there.

Spoiler:
You have no guarantee that the thing multiplying p is an integer. You want to multiply both sides by [imath]q^n[/imath] so you have integers all around. Then it follows that p divides [imath]a_0 \cdot q^n[/imath]. But p and q are relatively prime, so p divides [imath]a_0[/imath].

Oops >_<, you're right my proof doesn't make sense given the definition of divides. I'll rewrite it, I was just hoping to leave one trick for the reader to try and solve.
Last edited by polymer on Tue Feb 01, 2011 6:29 pm UTC, edited 1 time in total.

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