Proving that cube root 7 is irrational

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Discussion Overview

The discussion revolves around proving that the cube root of 7, \(\sqrt[3]{7}\), is irrational without relying on the unique factorization theorem. Participants explore various approaches and reasoning methods related to this proof, drawing parallels to the classic proof of the irrationality of \(\sqrt{2}\).

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests mimicking the proof for \(\sqrt{2}\) by assuming \(\sqrt[3]{7}\) is rational and expressing it as a fraction of integers.
  • Another participant elaborates on the initial approach, discussing the implications of cubing both sides and the necessity to show that if \(m^3\) is a multiple of 7, then \(m\) must also be a multiple of 7.
  • There is a mention of needing to consider different cases for numbers that are not multiples of 7 to explore the validity of the argument.
  • A later reply confirms the reasoning that if \(m^3 \equiv 0 \mod 7\), then \(m \equiv 0 \mod 7\), leading to a contradiction regarding the initial assumption of \(m/n\) being in lowest terms.

Areas of Agreement / Disagreement

Participants generally agree on the approach to proving the irrationality of \(\sqrt[3]{7}\) by using a method similar to that for \(\sqrt{2}\). However, the discussion includes varying levels of detail and clarity in the reasoning, indicating some uncertainty in the steps involved.

Contextual Notes

Some assumptions about the properties of numbers and modular arithmetic are present but not fully explored, leaving certain mathematical steps unresolved.

TalonStriker
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Hi guys,

How would you prove that \sqrt[3]{7} is irrational without using the unique factorization thrm? I tried proving that \sqrt[3]{7} is rational but it didn't seem to get me anywhere...

Thanks

EDIT: Looks like I posted this in the wrong forum.
 
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I would be inclined to mimic the classic Euclid proof that \sqrt{2} is irrational.

Assume that \sqrt{7} is rational. That is, assume \sqrt{7}= \frac{m}{n} for integers m and n, reduced to lowest terms. Then, cubing both sides, 7= \frac{m^3}{n^3} so 7n3= m3. That tells us that m3 is a multiple of 7. Can you use that to prove that m itself must be a multiple of 7? Remember that in proving that if m2 is even, the m must be even, we have to show that the square of an odd number is always odd. Here, you will have to look at numbers that are not multiples of 7. Is it possible for the third power of a number that is not a multiple of 7 to be a multiple of 7? You will have to look at 6 different cases.
 
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HallsofIvy said:
I would be inclined to mimic the classic Euclid proof that \sqrt{2} is irrational.

Assume that \sqrt{7} is rational. That is, assume \sqrt{7}= \frac{m}{n} for integers m and n, reduced to lowest terms. Then, cubing both sides, 7= \frac{m^3}{n^3} so 7n3= m3. That tells us that m3 is a multiple of 7. Can you use that to prove that m itself must be a multiple of 7? Remember that in profing that if m2 is even, the m must be even, we have to show that the square of an odd number is always odd. Here, you will have to look at numbers that are not multiples of 7. Is it possible for the third power of a number that is not a multiple of 7 to be a multiple of 7? You will have to look at 6 different cases.

Yes I can prove that m^{3} mod 7 = 0 implies mod 7 = 0. I assume that the next step would be set m=7k. Then plug it into the previous formula which would yield something like

7n^{3} = 7 * 49k^{3} using this I can prove that n % 7 = 0. So m/n has a common factor, which contracts one of the premises.

And that would be the proof, correct?
 
Yes, that was exactly what I meant.
 
thanks!
 

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