Question

proof that √2 is irrational
assuming that √2 = p/q, where p and q are integers not both even

Answer 1

winding up with 2q^2 = p^2
makes p^2 an even number

Answer 2

ok
so
2=(p/q)^2
so
2=p^2/q^2
2q^2=p^2
so we know p is even because p^2 is even
now
2q^2 = (2k)^2 = 4k^2
q^2 = 2k^2
so q is even
this is a contradiction to (p,q) = 1
thus
sqrt(2) is irrational

Answer 3

assume q^2 is even and q is odd
q^2 = (2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1 = odd
a contradction to q^2 is even
so
q^2 is even imples q is even

Answer 4

now please don't murder me for this

Answer 5

My favorite way of proving this statement is using the Rational Roots theorem.

Answer 6

Second favorite way is using the Fundamental Theorem of Arithmetic.

Answer 7

that's what I was trying to do next @joemath314159 but it seems tedious

Answer 8

Thats not very precise, can i see your proof anyway out of boredem

Answer 9

Which one, the Rational roots? or Fundamental Theorem? I think they are both shorter than the standard proof by contradiction.

Answer 10

the fundamental theorem 1

Answer 11

Sure. So the Fundamental theorem of Arithmetic is the one that says we can factor any integer uniquely into primes. So I'm going to prove \(\sqrt{2}\) is irrational by contradiction with that theorem.

Assume \(\sqrt{2}=\frac{a}{b}\) for some integers a and b, \(q\ne 0\). Then \(2b^2=a^2\). By the Fundamental Theorem of Arithmetic, we can factor a as \(a=p_1p_2\cdots p_n\) and b as \(b=q_1q_2\cdots q_m.\) Therefore:\[2b^2=a^2\Longrightarrow 2(q_1q_2\cdots q_m)^2=(p_1p_2\cdots p_n)^2.\]Now we count the number of primes on each side. There are \(2m+1\) primes of the left hand side (the +1 because of the 2 hanging out in the front), and \(2n\) on the right. Since m and n are natural numbers, an odd number can never equal an even. This is the contradiction.

Answer 12

er, that shouldnt be \(q\ne 0\), it should be \(b\ne 0\). my bad >.<

Answer 13

it's the same ending as @zzr0ck3r indicated

Answer 14

The parity of m and n themselves isn't relevant. Its the parity of 2m+1 and 2n.

Answer 15

Oh I see in oder for them to be equal they must have the same number of prime factors which would require that $$2m=2n+1$$

Answer 16

Exactly :)

Answer 17

p and q shouldn't have a factor in common?

Answer 18

yes

Answer 19

the contradiction winds up both p and q having he same factor of 2

Answer 20

Thats correct :)

Answer 21

I think I know what you did there, @joemath314159 but shouldn't you have assumed that p and q also have to be in lowest terms?

Answer 22

@Jack199 what is not percise?

Answer 23

With the F. T. A. approach, you aren't aiming to contradict anything dealing with reduced fraction \(\frac{p}{q}\). So it doesn't matter if the fraction is in lowest terms or not.

Answer 24

I wasn't responding to anything you wrote I was responding to joe, but then he clarified with a proof.

Answer 25

ahh. I think this proof (i think this is how one of the Pythagoreans proved it) is the best proof because it relies on nothing but knowledge of parity, and relatively primness.

Answer 26

Its a classic for sure.

Answer 27

I remember my proof in 4th grade went along the lines that any number in fractional form, if the quotient would seem to end up as non-terminating, it would fit the description of irrational if you attempted to finish it. LUL the person is the one who isn't rational ;)

Answer 28

The textbook standard for learning how to prove things by contradiction.

Answer 29

thanks both @joemath314159 and @zzr0ck3r

Answer 30

I think this and the proof that there are infinite primes, using the idea of dividing all the primes + 1 with a prime, are my favorite proofs

Answer 31

Indeed. Its so elegant.

Answer 32

I like things that are proved with the pigeonhole principle.

Answer 33

square root of any prime number is irrational?

Answer 34

yeah

Answer 35

Yes. In fact, the prove i presented using the F.T.A only works if we are dealing with the square root of the prime. The method zzr0ck3r presented works for any case.

Answer 36

proof*

Answer 37

if a*a = prime then a|p

Answer 38

what the heck

Answer 39

there are probably 20+ ways to prove it.

Answer 40

there are over 100 documented proofs for the pythagorean theorem
One of them was done by President Garfield and its actually one of the cooler ones.

Answer 41

mind boggled

Answer 42

the infinite primes one is like this

assume there are finite primes
{p,p_1,p_2....}
P = the product of them all + 1
P = p_1*p_2....p_n+1
so P is not prime because its the product of all the primes + 1
so P is divisible by some prime = p because P is composite
so p|P
but p cant divide our product of primes because then it would be in the product, so it must divide 1. A contradiction

I love this.

Answer 43

I appreciate the time both of you have poured into this seemingly "popular" and "classic" proof problem.