Question

$\sqrt{5 = \sqrt{17} }$
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Answer 1

Answer:

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Answer 2

$\huge\mathfrak\red{Answer :) }$

How do you prove that 17−−√5 is irrational?

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Carefully!

The fifth root of any whole number that isn’t a perfect fifth power is irrational.

First, let’s establish that 17 is not a perfect fifth power (hereafter referred to as a P5 for the sake of brevity). The prime factorization of any P5 must contain only prime factors with multiplicities of 5, or multiples of 5. Now that’s a lot to say, so let’s unpack it.

We know that 32 is a P5, because its prime factorization is 2⋅2⋅2⋅2⋅2 or 25 . The factor of 2 has a multiplicity of 5. For that matter, 1024 is also a P5, because its prime factorization is 210 , and 10 is a multiple of 5.

7776 ( 25⋅35 ) is a P5, but 23,328 ( 25⋅36 ) is not, because the multiplicity of 3 is 6, which is not divisible by 5.

Now back to 17. The prime factorization of 17 is…17. Because 17 is prime. So the multiplicity of 17 is 1, which isn’t divisible by 5; ergo, 17 isn’t a P5.

Now let’s prove the original statement: that the fifth root of any whole number that isn’t a P5 is irrational

Start by assuming the opposite. Assume that n−−√5=ab , where n is not a P5, and a and b are co-prime integers with b≠0 .

Raise both sides to the fifth power:

(n−−√5)5=n=a5b5

Now multiply both sides by b5 .

nb5=a5

Ah, but now we have a contradiction! Because n is not a P5, at least one of its prime factors has a multiplicity other than five; ergo, nb5 is also not a P5. But a5 is a P5. So we’re saying that a non-P5 is equal to a P5, which doesn’t make sense. Conclusion: the fifth root of a non-P5 is irrational.