Root(2) is irrational (Proof by contradiction!)

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[@pedroivog.s.6870]

Very straightforward :)

[@addwithad]

Great to hear :)

[@parkertolernew8520]

Video length has the same starting digits as square root of 2! (1:41)

[@addwithad]

Would you believe that I didn't even plan it haha

[@mitchtheronin1469]

Im a bit confused. Couldnt you do the same proof with 8?

[@plutoniumuser]

√8 is also irrational

[@plutoniumuser]

Sqrt(8) is also irrational

[@mitchtheronin1469]

@@plutoniumuser ok that was a coincidence 😅. Then lets take 4. i mean square root of 4 is obviously 2, but what would happen if you used the same method on a 4

[@sophiastern2719]

@@mitchtheronin1469 Lets look at sqrt(4)=a/b
4=a^2/b^2
a^2=4b^2 -> 4|a^2
however that only implies 2|a since a number like 6 which isn't divisible by 4 still squares into 36 which is
a=2k
(2k)^2=4b^2
4k^2=4b^2
k^2=b^2
Nothing can be determined about b's factors from the logic line since k is completely unknown, so no contradiction is possible.
In general the square root of any prime number is irrational, as is any odd power of that square root.
Additionally multiplying exponents of different prime roots behaves similarly to multiplying evens and odds, since the roots can only rationalize other roots of the same prime.
Multiplying exponents of different prime roots will result in an rational number if both of the inputs are rational, and an irrational number otherwise, so in order to get a rational number when square rooting a natural number, all of its prime factors need an even exponential, which happens exactly when the number is a perfect square.
Thus it is impossible to take the square root of a natural number and obtain a rational non-integer.

[@Vapor817]

​​@@mitchtheronin1469if you substitute sqrt of 2 with sqrt of 4 then the 4's end up cancelling out in the 4k^2 expression which leaves no contradiction as k^2 = b^2 in this scenario so b^2 can still be odd