Two funny (but unrelated) proofs

simX

simX

Unofficial Mac Genius
OK, I saw this first proof on the EV Nova boards a while ago, but I thought it'd be funny if I posted it here. Remember that this represents the views of one of the people on the EV Nova web boards, not MY views.

girls-are-evil.jpg


OK, and the second proof is that are number system is totally messed up. Here's the proof:

First, say we have two numbers a & b, and they are equal. Thus:

a=b

Multiply by a.

a^2 = ab

Subtract b^2.

a^2 - b^2 = ab - b^2

Factor.

(a+b)(a-b) = b(a-b)

Cancel the (a-b) factors.

a + b = b

Remember that a=b, thus we get

b + b = b

Group like terms.

2b = b

Divide by b.

2 = 1

Scary, huh? A corollary is that if two numbers are equal, both equal 0. Just go from

a + b = b

to

a = 0

by subtracting both sides by b.

Discuss.
 
LOL! I like the first proof best!

Second Proof: The number system is indeed messed up. I dunno what to say about this, except that you should show someone who is a professor this proof.
 
Sorry -- gotta kill this... ;)
a=b

Multiply by a.

a^2 = ab

Subtract b^2.

a^2 - b^2 = ab - b^2
Click to expand...
a = b

So when you multiply by a you get

a^2 = ab

but substitute a back in for b.

a^2 = a^2

subtract b^2 (which is really a^2)

a^2 - b^2 = a^2 - b^2

since b^2 = a^2 you get

0 = 0

end of faulty proof. :)
 
Originally posted by nkuvu
Sorry -- gotta kill this... ;)

a = b

So when you multiply by a you get

a^2 = ab

but substitute a back in for b.

a^2 = a^2

subtract b^2 (which is really a^2)

a^2 - b^2 = a^2 - b^2

since b^2 = a^2 you get

0 = 0

end of faulty proof. :)
Click to expand...

Um, no. You can't disprove a proof by providing an alternate ending! Where did YOU learn algebra? :p

Seriously, though, you can't disprove something like that. You have to find fault with the specific steps that I did.

Think of it this way, say I take 3 pencils and arrange them in the form of the letter A. Then you take them and arrange them in the form of the letter N. You can't say, "It's impossible to arrange them like an A, because one can arrange them like an N!"

Basically what your proof said is that 0 = 0. Of course! We all know that. My proof is much more interesting; 2 = 1, which we didn't all know.
 
2 = 1? So I have only five fingers? And macosx.com avatars have 25x25 pixels? Oh my god!:eek:

I feel a stream of eeks coming:

:eek::eek::eek::eek::eek::eek:
 
Originally posted by simX

Um, no. You can't disprove a proof by providing an alternate ending! Where did YOU learn algebra? :p

Seriously, though, you can't disprove something like that. You have to find fault with the specific steps that I did.
Click to expand...

Oh, all right.

Factor.

(a+b)(a-b) = b(a-b)
Click to expand...
Since a = b, a - b = 0. So when you divide through by (a - b), you're dividing through by zero, and you don't even need to know algebra to know that that's the kiss of death.

Me, I prefer:
Proof that Horses Have An Infinite Number of Legs (Not Remotely Valid)
Everyone knows that horses have two sets of legs, hind legs and forelegs. So they have two hind legs and then forelegs, but we also know that 2 legs + 4 legs = 6 legs. Now 6 is an even number, but certainly an odd number of legs for a horse to have. But the only number that is both even and odd is infinity. So horses have an infinite number of legs.

Whew, that one's bad.

Math jokes? Yeah, we got those.

-the valrus
 
Originally posted by Valrus


Oh, all right.


Since a = b, a - b = 0. So when you divide through by (a - b), you're dividing through by zero, and you don't even need to know algebra to know that that's the kiss of death.
Click to expand...

Now THAT'S the correct way to poke holes in a proof. I was waiting to see how long it would take before somebody noticed that. ;)
 
You see these all over the place, and invariably they involve dividing by zero. Whenver you want to blow a hole in these proofs, just look for the division step.

However they do introduce an interesting concept: different amounts of infinity.

: when you divide by 1/(a-b) you could substitute in @ (sorry - nearest I could find to infinity symbol) and you end up with something like @ = 2@.

That reminds me of this situation:

Consider the series of integers:
1,2,3,4............ it is an infinite series.

consider now the series of even numbers:
2, 4, 6, 8 .................... another infinite series

However there are twice as many numbers in the integer series as opposed to the even number series.

Can it be said that the integer series is more infinite than the even number series? Interesting situations like this are re-defining the way that we look at infinity.

Roger.
 
For all who cares the result for the first proof is not

[Girls=Evil]

It must be

[Girls<Evil]

Why? Because, as is stated "Girls require time AND money" is not

[Time X Money]

but

[Time + Money]

So the formula is

[Girls=Money+Money=2Money]

[Girls=2(?evil)]

So, finally

[Girls <evil]:D

Anyone for tennis?

(Björn
 
Originally posted by roger:
Can it be said that the integer series is more infinite than the even number series? Interesting situations like this are re-defining the way that we look at infinity.
Click to expand...

Erm, not really actually. The concept of different "levels" of infinity has been around for a while. There was a German named Georg Cantor in the 19th century who defined the different levels of infinity. As it pertains to integers, the set of all integers has the same number of elements as the set of even integers because there is a one-to-one correspondence between the two sets:
-1 -> -2
0 -> 0
1 -> 2
2 -> 4
And so on. So m, in the set of integers, corresponds to n, in the set of even integers such that n = 2m. In the same way, the set of integers divisible by any k is of the same size as the set of integers, via the correspondence n -> kn.
In fact, the set of rational numbers (fractions with integral numerators and denominators) is also of the same size, although the reasoning is a bit more involved. It's called the "diagonal argument" for anyone interested.
That level of infinity is called "aleph-null," after the first letter of the Hebrew alphabet. The real numbers, on the other hand, are actually in a larger order of infinity - aleph-one - than the integers (meaning they can't be put into a one-to-one correspondence with the integers), and the proof is really cool. I won't put it up here unless someone asks, though, since this is probably too long already. For that matter, I could put up the diagonal argument as well.

Hope someone else here finds all this as interesting as I do... :D

-the valrus
 
Perhaps if you said that "Girls = Money spent over a period of time", then you would have "Money x Time", and hence, girls are still evil.:D
 
Originally posted by Matrix Agent
Perhaps if you said that "Girls = Money spent over a period of time", then you would have "Money x Time", and hence, girls are still evil.:D
Click to expand...

This is getting closer, and perhaps we will find as we continue this that:

[Girls > Evil]

Dear mother of god! (Whoever she might be):confused:
 
Originally posted by kvist
For all who cares the result for the first proof is not

[Girls=Evil]

It must be

[Girls<Evil]

Why? Because, as is stated "Girls require time AND money" is not

[Time X Money]

but

[Time + Money]

So the formula is

[Girls=Money+Money=2Money]

[Girls=2(?evil)]

So, finally

[Girls <evil]:D

Anyone for tennis?

(Björn
Click to expand...

Well, in boolean algebra, AND is multiplication, OR is addition
0 AND 1 = 0 * 1 = 0
0 OR 1 = 0 + 1 = 1

So the sense of AND is not necessarily clear. In any case, it only clearly states a requirement of time and money, not a strict equality, so we are down to something unimpressive like

girls = A + B(time AND money)^C

with A, B, and C arbitrary expressions, and the meaning of AND unclear.

to return to your last step though

from
[Girls=2(?evil)]
to
[Girls <evil]

If evil = 4, then girls = 2(evil^0.5) = 2(2) = 4
If evil's less than 4 then girls > evil, if evil's greater than 4 then girls < evil
 
Originally posted by scruffy
If evil = 4, then girls = 2(evil^0.5) = 2(2) = 4
If evil's less than 4 then girls > evil, if evil's greater than 4 then girls < evil
Click to expand...

You mean that evil is a number? :confused:
 
Originally posted by xoot


You mean that evil is a number? :confused:
Click to expand...

...and is a Girl a boolean expression? What about true love? Is there a hidden message in this?:confused:

Perhaps girls really are antimatter, so [Girl+Boy=0]?

/Björn
 
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