Which proof is better?

[bunny]

Sqrt(2) is irrational.

Proof One:
Suppose Sqrt(2) = p/q where p and q are integers with no common factors.
implies 2 = p^2/q^2
implies 2q^2 = p^2
implies p^2 is even
implies p is even
implies p = 2k for some k
implies 2q^2 = (2k)^2
implies 2q^2 = 4k^2
implies q^2 = 2k^2
implies q is even
This contradicts original assumption, therefore Sqrt(2) cannot be written in the form p/q where p and q are integers with no common factors
Thus, Sqrt(2) is irrational.

Proof Two:
Suppose Sqrt(2) = p/q where p and q are integers with no common factors.
implies 2 = p^2/q^2
implies 2q^2 = p^2
since q = (a1^n1)(a2^n2)…..(ak^nk) for primes a1 to ak
and p = (b1^m1)(b2^m2)…..(bj^mj) for primes b1 to bj
this implies:
2(a1^2n1)(a2^2n2)…..(ak^2nk)=(b1^2m1)(b2^2m2)…..(b j^2mj)
This contradicts the fundamental theorem of arithmetic, since the prime decomposition of a number is unique and the power of 2 is odd on one side of this equation and even on the other.
Thus Sqrt(2) is irrational.

[madnak]

I say proof two. It's much more elegant and powerful.

[Borodog]

Proof one. Much simpler and clearer.

[jstnrgrs]

I understand proof 1 with no problem at all.

proof 2, I really have to think hard about to understand.

Therefore, I think proof 1 is better.

[SBR]

I voted neither, it probably depends on the context. I.e. proof 1 is much easier to understand therefore if you were teaching a high school class it would be "better". But the method used in proof 2 is more powerful.

[chezlaw]

p^2 is even implies p is even (in proof 1).) bothers me a bit.

Not cause its not true but p^2 is divisible by 23 therefore p is divisible by 23 wouldn't slip by so easily.

In proof 2) we get het up about p=q -> p and q have the same parity of the factor 2 which (keep dividing by two) amounts to much the same thing.

Anyway much prefer 2)

chez

[bunny]

[ QUOTE ]
p^2 is even implies p is even (in proof 1).) bothers me a bit.

Not cause its not true but p^2 is divisible by 23 therefore p is divisible by 23 wouldn't slip by so easily.

[/ QUOTE ]
Could probably solve your unease by referencing the theorem that odd numbers have odd squares and even numbers have even squares. There are a couple of unjustified steps in both but I think the gist is there.

[Borodog]

[ QUOTE ]
[ QUOTE ]
p^2 is even implies p is even (in proof 1).) bothers me a bit.

Not cause its not true but p^2 is divisible by 23 therefore p is divisible by 23 wouldn't slip by so easily.

[/ QUOTE ]
Could probably solve your unease by referencing the theorem that odd numbers have odd squares and even numbers have even squares. There are a couple of unjustified steps in both but I think the gist is there.

[/ QUOTE ]

Hmm. I took that to be intuitively obvious. Maybe it isn't.

[thylacine]

[ QUOTE ]
[ QUOTE ]
p^2 is even implies p is even (in proof 1).) bothers me a bit.

Not cause its not true but p^2 is divisible by 23 therefore p is divisible by 23 wouldn't slip by so easily.

[/ QUOTE ]
Could probably solve your unease by referencing the theorem that odd numbers have odd squares and even numbers have even squares. There are a couple of unjustified steps in both but I think the gist is there.

[/ QUOTE ]

Both proofs use that for p prime: if p divides ab then p divides a or p divides b.

[bunny]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
p^2 is even implies p is even (in proof 1).) bothers me a bit.

Not cause its not true but p^2 is divisible by 23 therefore p is divisible by 23 wouldn't slip by so easily.

[/ QUOTE ]
Could probably solve your unease by referencing the theorem that odd numbers have odd squares and even numbers have even squares. There are a couple of unjustified steps in both but I think the gist is there.

[/ QUOTE ]

Hmm. I took that to be intuitively obvious. Maybe it isn't.

[/ QUOTE ]
Ultimately it depends on the fundamental theorem of arithmetic in that it's a special case of "n^2 is divisible by prime p implies n is divisible by p also". Although I think the special case is intuitively obvious to people who have played around with numbers and arithmetic, I doubt the general case is, a la chezlaw's 23 example.