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        riddles >> medium >> any number equals any number
        
(Message started by: EZ_Lonny on Aug 11^th, 2004, 6:03am)

Title: any number equals any number
Post by EZ_Lonny on Aug 11^th, 2004, 6:03am

This is a golden oldy, but still fun.

Assume:
x = y then

x[smiley=sup2.gif] = y[smiley=sup2.gif]
x[smiley=sup2.gif] – y[smiley=sup2.gif] = y[smiley=sup2.gif] – y[smiley=sup2.gif]
x[smiley=sup2.gif] – y[smiley=sup2.gif] = 0
(x[smiley=sup2.gif] – y[smiley=sup2.gif]) / (x – y) = 0 / (x – y)
x + y = 0 remind that x = y, so
x + x = 0
2x = 0
2x / 2x = 0 / 2x
1 = 0

So any given number equals 0 and therefore equals any other number

Title: Re: any number equals any number
Post by Three Hands on Aug 11^th, 2004, 6:12am

I think the main problem with that sequence comes in the [hide]fourth[/hide] line:

::[hide]
Given that x=y, (x-y) = 0, hence what you are doing in this line is dividing by 0, which causes all sorts of problems in maths. In fact, it's not entirely certain what 0/0 equals (whether it is undefined (or infinity) or 0). In any case, the left hand side of the equation would equal some undefined large value, rather than 0[/hide]
::

Title: Re: any number equals any number
Post by towr on Aug 11^th, 2004, 9:19am

Starting with the assumption any two numbers (x,y) are the same does not seem to me a good way to prove any two numbers are equal..
You should either assume they're different and arrive and a contradiction, or assume nothing and prove they have to be equal.

Since you've started with asuming x=y, and arrive at 1=0 which is plainly false, you've proven no two numbers are the same.. :P

Title: Re: any number equals any number
Post by Icarus on Aug 11^th, 2004, 7:30pm

on 08/11/04 at 06:12:47, Three Hands wrote:

In fact, it's not entirely certain what 0/0 equals (whether it is undefined (or infinity) or 0).

No. It is entirely certain that 0/0 is undefined. We could of course define it to be any thing we like. But no matter what definition we could give it, the only result would be to add additional exceptions to statements involving divisions (instead of simply saying "except when the denominator would be zero", like we do now, we would have to say "except when the denominator would be zero and either the numerator is not zero or the other side is equal to ..."). I.e. the situation that results from giving a definition to 0/0 is even more complicated than the one where it is not defined. For this reason, we prefer to leave 0/0 undefined.

Title: Re: any number equals any number
Post by rmsgrey on Aug 12^th, 2004, 3:49am

Except when the 0/0 arises in contexts like that of a (locally) continuous function, like x/sin(x) which at x=0 has value 0/0=1, where the useful property of maintaining continuity outweighs the problems of defining 0/0 for that specific pair of 0's

Title: Re: any number equals any number
Post by EZ_Lonny on Aug 12^th, 2004, 8:41am

I think it could be a better set of equations if I made it with limits,

saying: lim x[smiley=subn.gif]->y;

With that there are some mistakes ruled out

Title: Re: any number equals any number
Post by rmsgrey on Aug 12^th, 2004, 2:51pm

So if you take the limit as x->y, you can represent that as taking y = x + h and taking the limit as h->0.

That means that instead of x² = y², you have:
x² + 2hx + h² = y[sup2[/sup]
x² - y² + 2hx +h² = y² - y²
x² - y² + 2hx +h² = 0
(x² - y²) / (x - y) + 2hx / (x - y) +h² / (x - y) = 0 / (x - y)
x + y + 2hx / (x - y) +h² / (x - y) = 0
using x - y = - h:
x + y - 2x - h = 0
using y - h = x:
x + x - 2x = 0

so 0 = 0 and I'm not the pope after all...

The key point is that, since you're dividing by h at one point, you can't neglect the order h terms - or to put it another way, since you're working with the difference between x² and y² later, you can't start by ignoring it...

Title: Re: any number equals any number
Post by Icarus on Aug 18^th, 2004, 7:34pm

on 08/12/04 at 03:49:17, rmsgrey wrote:

Continuous extension does not represent a definition for 0/0. The reason 0/0 is undefined is that a priori, there is no reason to choose any one possible value over the others. In particular situations - such as this one - one value is preferable. So when these situations occur, we note the choice of value. But different situations will prefer other values.

For this reason, the situation you describe is not normally expressed as "defining 0/0 = 1 in this situation", but rather, it is expressed as "extending continuously to 0". I.e., it is not 0/0 being defined, but the function f(x) which is defined elsewhere by x/sin x.