Proving that cube root 7 is irrational

In summary, the proof uses the fact that if m2 is even, the m must be even, to show that the square of an odd number is always odd. This in turn proves that m itself must be a multiple of 7.
  • #1
TalonStriker
15
0
Hi guys,

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

Thanks

EDIT: Looks like I posted this in the wrong forum.
 
Last edited:
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  • #2
I would be inclined to mimic the classic Euclid proof that [itex]\sqrt{2}[/itex] is irrational.

Assume that [itex]\sqrt{7}[/itex] is rational. That is, assume [itex]\sqrt{7}= \frac{m}{n}[/itex] for integers m and n, reduced to lowest terms. Then, cubing both sides, [itex]7= \frac{m^3}{n^3}[/itex] 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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  • #3
HallsofIvy said:
I would be inclined to mimic the classic Euclid proof that [itex]\sqrt{2}[/itex] is irrational.

Assume that [itex]\sqrt{7}[/itex] is rational. That is, assume [itex]\sqrt{7}= \frac{m}{n}[/itex] for integers m and n, reduced to lowest terms. Then, cubing both sides, [itex]7= \frac{m^3}{n^3}[/itex] 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[tex]^{3}[/tex] 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

[tex]7n^{3} = 7 * 49k^{3}[/tex] 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?
 
  • #4
Yes, that was exactly what I meant.
 
  • #5
thanks!
 

What does it mean for a number to be irrational?

A number is considered irrational if it cannot be expressed as a ratio of two integers (a fraction) and its decimal representation is non-terminating and non-repeating.

How do you prove that cube root 7 is irrational?

To prove that cube root 7 is irrational, we can use proof by contradiction. We assume that cube root 7 is a rational number, and then show that this assumption leads to a contradiction, thus proving that it is actually irrational.

Can you give an example of a proof by contradiction for cube root 7 being irrational?

Yes, we can assume that cube root 7 is rational and can be written as a fraction m/n, where m and n are integers with no common factors. Then we can cube both sides of the equation to get 7 = (m/n)^3, which simplifies to 7n^3 = m^3. This means that m^3 is divisible by 7, and therefore m must also be divisible by 7. However, this means that n^3 is also divisible by 7, which contradicts our initial assumption that m and n have no common factors. Therefore, cube root 7 cannot be rational.

Why is proving that cube root 7 is irrational important?

Proving that cube root 7 is irrational is important because it is an example of a larger concept in mathematics - the existence of irrational numbers. It also helps us understand the properties and behavior of irrational numbers, which are essential in many areas of mathematics and science.

Are there other ways to prove that cube root 7 is irrational?

Yes, there are other approaches to prove that cube root 7 is irrational, such as using the fundamental theorem of arithmetic or using the concept of prime factorization. However, the proof by contradiction is the most common and straightforward method to prove that cube root 7 is irrational.

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