prove that a rational number squared has each of its prime factors with even exponent


by mccalli1
Tags: exponent, factors, number, prime, prove, rational, squared
mccalli1
mccalli1 is offline
#1
Oct4-12, 07:07 PM
P: 5
I have a theory that i need to prove but im not quite sure how to mathematically prove that it is true.

Theory: When you square a rational number, each of the prime factors has an even exponent.

For example,

10 --> If i square 10, which is a rational number,
=10^2
=(5^2 x 2^2) --> both 5 and 2 are prime, and have even exponents. Thus, 10 is a rational number.

√7 ?
=(√7)^2
=7^1 --> odd exponent, thus irrational number.

I want to prove this will work for any case. Any ideas??
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A. Bahat
A. Bahat is offline
#2
Oct4-12, 07:25 PM
P: 150
What you do consider to be the prime factors of, say, 1/2?
Eval
Eval is offline
#3
Oct4-12, 07:30 PM
P: 33
You know that all numbers can be expressed as a product of primes and 1. Then, if you write your number n as n=p1c1p2c2p3c3... (a general prime factoring), what happens when you square this number? (Hint: The "p" values are not the important part.)

Once you have found this, what can you say about a number that has an odd power for a prime factor?

Finally, what do you know about multiplying rational and irrational numbers together?

I hope I didn't give too much away!

Alcatrace IV
Alcatrace IV is offline
#4
Oct4-12, 10:28 PM
P: 9

prove that a rational number squared has each of its prime factors with even exponent


I infer that you are a mathematical amateur (like me, a highschool freshman - perhaps we can personally relate?) and are in requirement of guidance.

Umm ... a little advice, a small tip of insurance. For any composite number 'n' raised to any power 'p', the prime factorization of it yields the base coefficients of 'n' raised to the same power 'p'. For example, 15^2 is the same as 5^2 multiplied by 3^2. Also, 12^3 is equivalent to 3^3 X (4^3 = 2^6 < 2^3X2^3) etc. An implication of this is that for every p-powered number ... what law do the patterns logically entail?

I think you can complete the paragraph by filling in the blanks. Not intending to spoil the fun, I've refrained from teasing the horses lest they run away from the stable. On that note, I'm fairly new around this place as well and wish you luck!
mccalli1
mccalli1 is offline
#5
Oct5-12, 07:56 AM
P: 5
Thank you all for your replies,

Eval, if i square my number "n" my prime factors = p^c1, p^c2, p^c3 ... will all be raised to the power of 2 as well. That part I got.

I know that a prime factor raised to a odd power should be an irrational number, after the number "n" has been squared. I am just having trouble proving it.

If you have a proof i would love to see it!! Thanks again
Eval
Eval is offline
#6
Oct5-12, 08:23 AM
P: 33
Basically, it works like this. Assume n=p1c1p2c2p3c3... Now, divide by all the primes with odd power. For example, if c3 is odd, divide by p3. now you have a product of primes to the first power multiplied by the product of primes to even power. the second product, then, is a perfect square, so you can divide all the powers in half and you still have an integer. The first number in the product, though, you have to show is not a perfect square. I think you can do this (you had the basic idea in your first post). When you have done this, you will show that the square root of n is an irrational times a rational, so it must be irrational.

This is not a rigorous proof, by any means since I am assuming that the square root of a non-square number is irrational and that an irrational times a rational is irrational. However, the proofs for these are also rather "simple." Unless you are assuming these for a class, most mathematicians will accept those assumptions because they have been proven many times. (Any of my professors would expect a complete, rigorous proof for a final draft, but in a discussion, they would find this acceptable.)

EDIT: Hmm, or are you looking for a rigorous proof that the square root of a non-square number is irrational? Now that I am rereading your first post, it seems like this is what you could be looking for. I was thinking along different lines. It seems like your topic title does not follow from your first post, so now I am confused.
mccalli1
mccalli1 is offline
#7
Oct5-12, 05:15 PM
P: 5
Thanks for the explanation but unfortunately it isnt what im looking for. Let me try to explain it clearer.

Say you have a number (n), any number and you want to know if it is rational or irrational. My theory is that if you square this number (n), then reduce it to its prime factors, those prime factors must have even exponents in order for n to be rational.If the exponents are odd, this means it is irrational.

I know this theory is true, however like you said i need a rigorous proof. Let me know if you come up with something!
acabus
acabus is offline
#8
Oct5-12, 05:29 PM
P: 45
Quote Quote by mccalli1 View Post
Thanks for the explanation but unfortunately it isnt what im looking for. Let me try to explain it clearer.

Say you have a number (n), any number and you want to know if it is rational or irrational. My theory is that if you square this number (n), then reduce it to its prime factors, those prime factors must have even exponents in order for n to be rational.If the exponents are odd, this means it is irrational.

I know this theory is true, however like you said i need a rigorous proof. Let me know if you come up with something!
So how does this work with [itex]\sqrt{18}[/itex], [itex]\sqrt[3]{7}[/itex], or [itex]\pi[/itex]? I'm just trying to show that you may be overlooking some possibilities.
mccalli1
mccalli1 is offline
#9
Oct5-12, 05:40 PM
P: 5
Sorry, i guess the correct statement would be any whole number squared..

18^1/2 --> square this
= 18
= 9x2
= 3x3x2 --> all odd exponents

7^1/3 --> squaring this would result a fraction in the exponent.

thus this proof wouldnt be aplicable as a fraction is nor even nor odd.

pi cannot be reduced to prime factors and also wouldnt be applicable.

thanks for keeping me on my toes
Eval
Eval is offline
#10
Oct5-12, 09:52 PM
P: 33
EDIT: This proof is flawed (thanks Norwegian!). Hopefully this works better :)

There is a generic proof for showing that the square root of a non-square integer is irrational. Let √k be your number where k is not square. Assume it is rational. That is, let [itex]\sqrt k=\frac{m}{n}[/itex] where [itex]m,n\in Z[/itex], [itex]gcd(m,n)=1[/itex] and [itex]n\neq 0[/itex]. Then we have that [itex]k=\frac{m^{2}}{n^{2}}[/itex] and thus [itex]kn^{2}=m^{2}[/itex]. Now, since √k is not an integer, n is not 1, and since gcd(n,m)=1, then k|m2. But since k is not a perfect square, this means k|m, so m=ak for some natural number a. Thus, k=a2k2/n2. Then we have that n2=a2k, but wait! This means that a|n, but a|m so gcd(m,n)≠1. This means we have reached a contradiction, and so this means that the square root of a non-square integer cannot be rational. Thus, it must be irrational. ∴

Let us test this with an example of √18. Let √18=m/n with the appropriate conditions. Then 18=m2/n2, 18n2=m2. Then 18|m2 and since 18 is not square, this means 18|m. so m=18a, and then 18=182a2/n2. Then, n2=18a2, so a|n, but a|m, so gcd(n,m)≠1, contradicting the assumption that √18 is rational, therefore it is irrational.

:)
Norwegian
Norwegian is offline
#11
Oct5-12, 10:41 PM
P: 144
Quote Quote by Eval View Post
Then 18|m2 and since 18 is not square, this means 18|m.
And how do you deal with the case m=6?
Eval
Eval is offline
#12
Oct6-12, 06:06 AM
P: 33
Touché. I'll think about where my proof went wrong.
EDIT: Since 18n2=m2, m2≥18. In fact, I showed that n≠1 when k is non-square, so m≥72. I get your point though.
EDIT2: Then what I should say is that since 18|m2, gcd(m,18)≠1. Here is a revised proof.

Let √k be a natural number where k is not square. Assume it is rational. That is, let [itex]\sqrt k=\frac{m}{n}[/itex] where [itex]m,n\in Z[/itex], [itex]gcd(m,n)=1[/itex] and [itex]n\neq 0[/itex]. Then we have that [itex]k=\frac{m^{2}}{n^{2}}[/itex] and thus [itex]kn^{2}=m^{2}[/itex]. Now, since √k is not an integer, n is not 1, and since gcd(n,m)=1, then k|m2. Then we know that gcd(k,m)≠1, so where k=pr, and gcd(k,m)=p, then p|m. So m=ap for some natural number a. Thus, k=a2p2/n2. Then we have that n2=a2p2/k=a2p/r. This means rn2=a2p. Since a and r are relatively prime (see Below), gcd(n,a)≠1 for a≠1. This means gcd(m,n)≠1, contradicting our assumption that k is rational, so let us look at the case where a=1. Then, rn2=p. Since k=pr, k=n2r2=(nr)2, contradicting our assumption that k is non-square.

Therefore, √k is irrational when k is a non-square natural number.

Below:
Otherwisw, gcd(k,m) would include a factor of r, and so it would be precisely k, but that would mean k|m and so n=1, but this would imply k is a perfect square
cheahchungyin
cheahchungyin is offline
#13
Oct6-12, 06:41 AM
P: 52
I think I know what you meant from the beginning. Here's some obvious facts.

1. Product of all prime factors is the number itself.
2. If A=B, then A^2 must equal B^2
3. From statement 1 and 2, We know that products of all prime factors squared is going to be the number squared.
4. Its obvious that all full numbers square is going to be the product of prime factors square.
mccalli1
mccalli1 is offline
#14
Oct6-12, 12:09 PM
P: 5
Quote Quote by Eval View Post
since 18|m2,
I see you use this notation but what does it mean?
Eval
Eval is offline
#15
Oct8-12, 11:59 AM
P: 33
Sorry for the wait. 18|m2 reads as "18 divides m squared." Basically, m2 is a multiple of 18, so it can be written as 18 times an integer.


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