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Prove 7th root of 7 is irrational

  • Thread starter Firepanda
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  • #1
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1st I assume it is rational so:

7^(1/7) = m/n

then

7 = (m^7)/(n^7)

implies m^7 is a multiple of 7.

Means m^7 = 0 mod 7

So, using fermats little theorem..

m^7 = m mod 7

for m to be in the class of 0 it has to be a multiple of 7.

Now set m = 7k, so

7n^7 = 49k^7

But now I'm stuck, how do I show from here that m/n has a common factor?

Thanks
 

Answers and Replies

  • #2
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Oh i see,

so carrying on..

7n^7 = 49k^7

implies that n^7 is a multiple of 7, and i can apply the same proof to show n is a multiple of 7 aswell using fermats little theorem.

So m, n have a common factor of 7. Which show's that m/n wasn't in its simplified form, which proves it by contradiction.

Correct? :)
 
  • #3
exk
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yes. You don't really need to invoke fermats little theorem. [itex]7=\frac{n^{7}}{m^{7}}[/itex] so [itex]7m^{7}=n^{7}[/itex] which means that m is a divisor of n (division algorithm), which contradicts your assumption.
 
  • #4
Dick
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7 is prime. If m^7 is divisible by 7 then m is divisible by 7. You don't need a fermat theorem and I don't think you need a division algorithm either.
 
  • #5
matt grime
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And 7^7 isn't 49.
 
  • #6
exk
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7 is prime. If m^7 is divisible by 7 then m is divisible by 7. You don't need a fermat theorem and I don't think you need a division algorithm either.
m being divisible by 7 doesn't contradict his assumption since both m and n need to be divisible by 7.
 
  • #7
Dick
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m being divisible by 7 doesn't contradict his assumption since both m and n need to be divisible by 7.
If m is divisible by 7 that means m^7 has at least 7 factors of 7. That leads pretty directly to the conclusion n is divisible by 7.
 
  • #8
exk
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If m is divisible by 7 that means m^7 has at least 7 factors of 7. That leads pretty directly to the conclusion n is divisible by 7.
I am sorry, I don't follow your thought there. He wanted to prove that 7 is irrational so he made it rational and said it can be expressed as [itex]7=\frac{m}{n}[/itex] and that the quotient is in lowest terms, i.e. m & n are coprime.

You say that because 7 is prime then m has to be divisible by 7. However, m being divisible by 7 doesn't make m coprime to n as far as I can see, but we need it to be coprime in order to get the contradiction. Could you please elaborate how that works out?
 
  • #9
Dick
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I am sorry, I don't follow your thought there. He wanted to prove that 7 is irrational so he made it rational and said it can be expressed as [itex]7=\frac{m}{n}[/itex] and that the quotient is in lowest terms, i.e. m & n are coprime.

You say that because 7 is prime then m has to be divisible by 7. However, m being divisible by 7 doesn't make m coprime to n as far as I can see, but we need it to be coprime in order to get the contradiction. Could you please elaborate how that works out?
7=m^7/n^7. Assume m and n are coprime. m^7=7*n^7 -> 7 divides m^7 -> 7 divides m -> m=7*k. So 7*n^7=(7^7)*k^7 -> n^7=(7^6)*k^7. Therefore 7 divides n^7. Therefore 7 divides n. Beep. (That's the 'contradiction' beep).
 
  • #10
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I used fermats theorem to prove that m was a multiple of 7 :)

I didn't know any other way to prove it, and I didn't just wnat to asume it without backing it up.
 
  • #11
Dick
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I used fermats theorem to prove that m was a multiple of 7 :)

I didn't know any other way to prove it, and I didn't just wnat to asume it without backing it up.
Just think of prime factorization, if a prime divides a product of numbers, then it must divide one of the numbers.
 

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