Which proof is better?

[bigpooch]

I would think Proof One is better because:

a) It's clearer; any intelligent person can folow it
b) It's concise; it's shorter
c) There are fewer algebraic symbols
d) The theorems or propositions that one needs to follow it
are simpler

If you could only pick one proof to show sqrt(2) is
irrational, which one would you pick?

On the other hand, if you wanted a "model proof" to show
that if n is not a square, sqrt(n) is irrational, then Proof
Two is more didactic.

Sometimes there are proofs that are much more complicated,
but shed light on something truly amazing or beautiful.
Most people are familiar with Euclid's proof of the
infinitude of primes, which is easy to follow. On the other
hand, a very elegant topological proof was found by
Furstenberg less than 50 years ago. This can be found at:

http://www.cut-the-knot.org/proofs/Furstenberg.shtml

That is just one example of a proof that many mathematicians
find interesting or elegant.

[evank15]

Not commenting on which one is better, but the argument used in proof 1 is known as "reductio ad absurdum", the most common method I've seen for proving root2 is irrational.

[CityFan]

Proof two depends on the fundamental theorem of arithmetic, which makes it messy an inaccessible.

In fact, proof one also depends on the fundamental theorem of arithmetic (p^2 is even
implies p is even) but I prefer to be spared the gory details.

[madnak]

[ QUOTE ]
b) It's concise; it's shorter

[/ QUOTE ]

Look again - P2 is shorter. And simpler, for that matter. It's definitely more abstract - I don't see that as a weakness.

[FortunaMaximus]

Neither. They end up saying the same thing.

[bigpooch]

Really? Depends on how you measure how long something is.
I use the number of keystrokes or characters.

Let's examine the actual text of the OP for the differences
(okay, the last line of the actual text of Proof One has an
extra comma, but materially does not differ in meaning from
the last line of Proof Two):

Proof One:

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


Proof Two:

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.



Let me use keystrokes (okay, I might be off a little, since
I didn't double check!).

The keystrokes (including an end-of-line, which I take as a
delimiter for the text) for Proof One is around 283. For
Proof Two, it is around 365.

Which do you think is longer? I suppose we could put a
proof in one line of text, but it would not necessarily be
"shorter".

Sorry for being such a "super-nit"! [img]/images/graemlins/smile.gif[/img]

[wazz]

Uh, don't they come to the same conclusion via the same lines? They are both 'reductio ad absurdiam' and they both rely on the fundamental theorem of arithmetic. I guess it's a question of which reference more external theories, so in that sense proof 2 is better. I hardly see how proof 2 is 'more powerful'.

[CityFan]

[ QUOTE ]
I guess it's a question of which reference more external theories, so in that sense proof 2 is better.

[/ QUOTE ]

You mean 2 is better because it DOES reference more external theorems, or because it doesn't?

Surely a self-contained proof is better, unless it is more complicated?

[bigpooch]

ONE
---

I would like to add that Proof One primarily uses two ideas:

[ Here, we take x,y, ... as integers. ]

The definiton of an even number.

a) An even number z is, by definition, divisible by two, so
there exists some integer k for which z = 2k. Thus, for any
integer j, the integer 2j is even. Also, by definition, an
even number has a factor of two.

Properties of square numbers. The property that is used
does NOT require the fundamental theorem of arithmetic.

b) If x^2 is even where x is an integer, then x is even.

Now, b) is equivalent to the contrapositive:

(Note that every integer is either odd or even: if one
divides an integer by two, the remainder is either zero or
nonzero)

b*) If x is not even, then x^2 is not even

or equivalently,

b**) If x is odd, then x^2 is odd.

This follows from the definition of an odd number: x is odd
implies there exists an integer m such that x = 2m+1. Also,
if n is any integer, 2n+1 is odd.

Proof of b**):

If x is odd, there is some integer m such that x = 2m+1.
x^2 = (2m+1)^2 = 4m^2 + 4m + 1 = 2(2m^2 + 2m) + 1.
Thus, x^2 is odd.

TWO
---

Proof Two uses the Fundamental Theorem of Arithmetic, and
although I agree that Proof Two is quite elegant, it still
boils down to the definition of odd and even numbers
(consider the exponents). So, essentially, Proof Two only
uses two ideas: a theorem and the definition of an even
number. Of course, one of the ideas (the theorem), is not
necessarily obvious to a layperson.


BETTER?
-------

It is somewhat subjective, and those that chose Proof Two
may have been swayed by the "aesthetics", and if that were
the only (or primary) criteria for choosing which is better,
then I would have chosen Proof Two. The primary reason for
choosing Proof One is that it does not rely on more complex
results and that was what I was trying to convey in my
previous post on this thread (e.g., proofs of the
infinitude of primes).

I think Proof One is better for the layperson or somebody
who wants to prove results from more basic or obvious
results.

Proof Two is better to illustrate how powerful a theorem can
be to prove other results or to give an elegant proof.

[madnak]

The text is irrelevant. Either proof could be expressed very gracefully, given the appropriate cypher. In fact, relatively basic symbols would suffice to render each of these proofs down to <100 keystrokes.

Proof One has more steps, however.