5= 4 !?!

Iancar29

OK lads....

PLEASE TELL ME WHERE I WENT WRONG HERE...

But... IMO i just proved something impossible? haha

OK... here we go ..

A=1, B=5, C=4
A = B - C
A(B - C) = (B - C)²
AB - AC = B² - 2BC + C²
BC +AB - AC = B² - BC + C²
BC + AB - B² = AC - BC +C²
B(C + A - = C( A + C -
Therefore: B = C
5 = 4 .... !?!?! :eek: :eek:

TheBody

Iancar29 wrote: »

OK lads....

PLEASE TELL ME WHERE I WENT WRONG HERE...

But... IMO i just proved something impossible? haha

OK... here we go ..

A=1, B=5, C=4
A = B - C
A(B - C) = (B - C)²
AB - AC = B² - 2BC + C²
BC +AB - AC = B² - BC + C²
BC + AB - B² = AC - BC +C²
B(C + A - = C( A + C -
Therefore: B = C
5 = 4 .... !?!?! :eek: :eek:

Ah the old classic. The problem is where you divided both sides by (C+A-B). (c+A-B) is 0, so you divided both sides by 0, which you can't do.

Iancar29

Cheers! ha... i never thought about subbing back in..

TheBody

It's a classic error. Just google "prove 2=1" and see all the pages that it throws up.

bnt

These things always seem to hinge on the Zeros: a zero is a black hole in the logic, the place where the trail goes cold, though it's not always obvious.
n x 0 = 0 for any value of n, or n / 0 is infinity, so the real value of n is lost whenever it encounters a zero, and anything goes. :cool:

Michael Collins

To be strictly correct division by zero is undefined, not plus or minus infinity - this confusion probably comes from one definition of how unbounded limits are treated

[latex] \displaystyle \lim_{x \to 0} \frac{1}{x} = \infty [/latex].

The reason it is undefined is easy to see. Clearly there's nothing wrong with

[latex] \displaystyle 0 \cdot 1 = 0 \cdot 2[/latex]

or

[latex] \displaystyle 0 \cdot 5 = 0 \cdot 4[/latex].

So mutiplication by zero, removes information as bnt says - or if you like, it is a many-to-one operation, taking any real or complex number to zero. As such, it doesn't have an inverse (since all real numbers are taken to zero, if you try to return from zero, how do you know which number to return to? It could have been any real number...)

Lets try to do it anyway

[latex] \displaystyle \frac{0 \cdot 1}{0} = \frac{0 \cdot 2}{0}[/latex]

[latex] \displaystyle \frac{0}{0} \cdot 1 = \frac{0}{0} \cdot 2 [/latex]

[latex] \displaystyle 1 = 2 [/latex]

Clearly not true. And those calculations may have looked a bit odd - the trouble is when zero is disguised as a variable, say [latex] x [/latex]

[latex] \displaystyle x \cdot 1 = x \cdot 2[/latex] *

Now try to simplify

[latex] \displaystyle \frac{x \cdot 1}{x} = \frac{x \cdot 2}{x}[/latex]

[latex] \displaystyle \frac{x}{x} \cdot 1 = \frac{x}{x} \cdot 2 [/latex]

[latex] \displaystyle 1 = 2 [/latex]

maybe not quite so easy to spot the error this time. Of course it is obvious that [latex] x [/latex] must be zero from * above, but often it is not so obvious, like in the first post.

There do exist structures where division by zero is allowed, such as using the extended set of real numbers

[latex] \bf{R} \cup \{\infty\}[/latex],

but this is not an algebraic field. Check out http://en.wikipedia.org/wiki/Real_projective_line