Proof that the square root of ANY integer is irrational (besides perfect squares)

Forget proving that the square root of 2 is irrational. We're proving that the square root of every positive integer is irrational, as long as it's not a perfect square!

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[angelmendez-rivera]

The core idea in this proof is that the standard form of the rational number p/q is defined such that gcd(p, q) = 1. One must know that if gcd(p, q) = 1, then gcd(p^2, q^2) = 1, and therefore, if p/q is not an integer, then neither is p^2/q^2. The reason gcd(p^2, q^2) = 1 is a consequence of gcd(p, q) = 1 is because gcd(p, q) = 1 implies that if sone prime factor a divides p, then it does not divide q. p^2 is just p multiplied by p, so the prime factorization of p^2 includes the factor a^2. However, if a does not divide q, then neither does a^2. If a^2 does not divide q, then it also does not divide q^2. Since this is true for an arbitrary prime a, it is true for all primes the prime factorization of p. Therefore, no prime factor of p^2 divides q^2. Therefore, gcd(p^2, q^2) = 1.

In general, the same proof is true for the kth root of n. This is because, analogously, gcd(p, q) = 1 implies gcd(p^k, q^k) = 1. This can be proven by induction on k, and the base case was already proven here. Therefore, if p/q is not an integer, then (p/q)^k for any integer for positive whole k is never an integer either. Hence, if there is no integer m such that m^k = n, then n^(1/k) is not rational.

[nitroh]

Really good logical proof, well done!

[samaiatraforti]

You're really great at explaining concepts! I am so grateful I found your channel

[hamidbennani]

Awesome job man, we won't have to worry about proofing √2, √3, ... are irrationals
Just like you said 👌🏻

[Fcalysson]

I've been looking for a proof of this for a long time, thank you so much

[hamzasiddiqui]

Wonderful video! I had to concentrate to understand the part where you concluded how n must not be an integer, I'll have to rewatch it to understand completely and root it in my head, but it was a very good explanation! All other steps were nicely explained as well, and the examples you gave were helpful in understanding them. Thank you for making this video!! :D

[okehjoseph]

wow...very relax and very precise ..nice work champ, nice soundtrack

[nychan]

This proof is even simpler than the proof of √2 is irrational in textbooks.

[goodplacetostop]

This is brillant, I love this proof. I can’t believe YouTube algorithms didn’t show me this channel before.

[hc_]

You can also go to the same process to show that the k-root of a non perfect k power is irrational as well

[MyOneFiftiethOfADollar]

Also cubed root of any natural number other than a perfect cube, Also fourth root of any natural number other than a perfect fourth power, is irrational

[noway]

Why no "Element of Q" or "Z" ? That's true gamer mathematics.

[francopascual]

Thank you so much man, really clean explanation

[crosseyedcat]

You can extend this to all rationals in canonical form by noticing that if p^2 / q^2 implies that every square root is going to only have half the prime factors in p and q. That means if the rational is in canonical form (gcd(p, q) = 1), then if either p or q has a prime factorization where one factor is repeated an odd number of times (i.e. it's not a perfect square), then it must be irrational. Using properties of the gcd, we can see if that if gcd(p, q) = 1, then gcd(p^2, q^2) = 1. So we can see that the new numerator in canonical form is going to just inherit the factors from p, but since squaring ensures all prime factors have an even multiplicity, there's no possible way that we can square p or q to get a rational number where either p or q has a prime factorization containing factors with an odd multiplicity. You can keep going in fact to show this for nth roots where if p or q have a prime factorization where a factor has multiplicity not divisible by n, it must be irrational.

[jonahansen]

It's easier for me to understand this proof than the standard one for the sqrt(2).

[otakurocklee]

Nice. I'd do it a bit differently. Assume p/q is in lowest terms. Hence they have no prime factors in common. q^2n = p^2. Assume there's a prime r that divides q. Then we know that r divides the left side. So r divides p^2, and hence r divides p (theorem that says if a prime s divides ab, then s|a or s|b). Therefore p and q have a prime factor in common which contradicts the assumption that p/q was in lowest terms. Hence there is no prime that divides q. So q=1. Which gives n=p^2. Which shows that if sqrt(n) is rational it must be a perfect square.

[kausheyaroy]

Is this proof correct ??
n = p^2/q^2 or nq^2 = p^2
So, q^2 is a factor of p^2
So, q is a factor of p
So, HCF (q, p) = q
But HCF (q, p) = 1 [acc. to our supposition]
It contradicts...
So, root n is irrational...

[Sir.raunakx]

This will help i my boards for 10th
Love from INDIA ❤❤

[rory]

thank you for helping me cram an assessment 15 hours before handin

[fmarten]

omg thats cool boy i finally understood it bruh
thnks buddy iam your new subscriber
literally amazing vidwo