Sqrt 2 is irrational proof

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

[JaybeePenaflor]

Every mathematics major needs to learn this proof at some point. I first learned this in High School, but only appreciated it when I was taking my mathematical proofs class.

[Harry-sfld]

a beautiful introduction to the proof by contradiction

[PrudentialViews]

There is a bit of history behind this. So basically a guy named Hippasus proved sqrt2 to be irrational. There is also another dude named Pythagoras who died believing that all numbers are rational. The group called "Pythagoreanism" lived on having the same belief. Hippasus was a Pythagorean and proved that sqrt2 was irrational and the other group members didn't like it so they threw him overboard and eventually killing him.

[Pink_queenofficial]

"911, what's your emergency?"
"My brain is not braining"

[shahmeer_haq]

I am in 10th Standard and our teachers have taught this proof in the class well because of your video i have now better understanding regarding this proof...Thank you so much 😀

[nishantbhardwaj]

My teacher taught me this in school, but after 6 years, finally i understood it. 😅

[himaneeshbhattacharya]

I learnt this is the seventh grade and haven't revised since. 4 yrs later, it's nostalgia

[raedentan]

The amount of people saying this can be used to prove √4 is irrational-
All your flawed proofs show is that m is even, not that m is a multiple of 4. Thus you cannot assume m = 4k and k is an integer. Also steamroller82 has been copy-pasting the same message over and over again trying to show them that they're wrong, really appreciate it.

[Timalsinaprithvi]

I am gonna use this method to prove that all the non-trivial zeros of the zeta function lie on the complex line whose real part is 1/2 and win myself a million dollars and an immortality in mathematics.

[varshinilolla]

I've first seen this in my class 10th math state book, and it still makes me feel amazed despite writing the same proof for a hundred times in my exams!!

[soupy]

Just to contribute another method that works for all prime integers:

There's a nice theorem that says for any polynomial in integer multiples, any Rational solution must be equivalent to a ratio of factors of the constant term and lead term.

To use this, consider the roots of
X^2 - P = 0 for any prime P

By definition, the only factors of P are 1 and itself, and since the lead term is 1, the only possible rational solutions are plus or minus 1 and P, but plugging these in doesnt work, so no root of P is rational.

Note 2 is prime, so it must have an irrational root.

This also works for higher roots then squares: just change the 2 in the exponent to whatever N you need.

[Maulik_Hadiyal]

Our sir taught that n/m is co-prime, since both are divisible by 2, thay are not co-prime, contradicting our assumption

[sharkhorse]

Took this on my foundations in computer engineering course. Very cool

[onso]

I remember it, when it was taught me in class 10 in india,

[Atomic]

damn this is a very neat and simple proof

[adwz]

I think a much better proof is to count the number of prime factors, this can prove sqrt(q) is irrational for all non-square integers q

Assume √q = a/b

==> a^2 = q b^2

q not square ==> there exists some prime power p in the decomposition of q of odd power

Now count number of prime factors p of q: a^2 has even power of p, b^2 has even power of p ==> qb^2 has odd power of p in decomposition (even + odd = odd)

Hence, a^2 =/= qb^2 is our contradiction, and q is irrational

So yeah, the square roots of 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, … are all irrational by the simple proof :)

[BALUDENDAGE]

Thankyou so much. First time when i saw it by another method i didnt understood but now i understand it with proof as well...thanks 😊

[maxfunpro]

I still remember this proof from my 9th class math lecture

[pNsB]

I also like the proof that if m and n are expressed as the product of their prime factors, then their respective powers of two will be different because one is even and one is odd.

[yungifez]

Proof by contradiction

Gotta love it