Square roots and such

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mcmesher
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Square roots and such

Postby mcmesher » Thu Jan 27, 2011 2:54 am UTC

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?
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mdyrud
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Re: Square roots and such

Postby mdyrud » Thu Jan 27, 2011 3:06 am UTC

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

Postby mcmesher » Thu Jan 27, 2011 3:11 am UTC

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.
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jestingrabbit
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Re: Square roots and such

Postby jestingrabbit » Thu Jan 27, 2011 3:14 am UTC

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

Postby mcmesher » Thu Jan 27, 2011 3:35 am UTC

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.
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jestingrabbit
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Re: Square roots and such

Postby jestingrabbit » Thu Jan 27, 2011 3:48 am UTC

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

Postby mdyrud » Thu Jan 27, 2011 4:03 am UTC

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

Postby WhiteRAZOR » Mon Jan 31, 2011 11:08 am UTC

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".

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Re: Square roots and such

Postby Yakk » Mon Jan 31, 2011 1:47 pm UTC

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
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Re: Square roots and such

Postby polymer » Tue Feb 01, 2011 8:28 am UTC

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].
[math]\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}[/math]


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.

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Qaanol
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Re: Square roots and such

Postby Qaanol » Tue Feb 01, 2011 10:01 am UTC

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].
[math]\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}[/math]

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

Postby mike-l » Tue Feb 01, 2011 4:38 pm UTC

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

Postby Entropyst » Tue Feb 01, 2011 5:11 pm UTC

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

Postby polymer » Tue Feb 01, 2011 6:03 pm UTC

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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