1 = 0

ozt9vdujny3srf

seeing as someone else jsut did something simialr, i may aswell just try my one (and yes, i know it is wrong, i just don't know why)

Its on the attachment (so that the superscript works)

Sev

1! = 0!
=> 1 = 0

Maths is only rules.

Lorax

1 is just a constant
use another number like 6 for example,
6^0 does not equal 6^1


my attempt anyway!

ozt9vdujny3srf

you can't cancel a factorial afaik

ecksor

Originally posted by Sev
1! = 0!

Only by definition. I.e, it doesn't follow logically from any axioms.

=> 1 = 0

That doesn't follow.

Maths is only rules.

But you're not following them ...

Bard

If x^n = x^b then n = b ?

Yeah, sure ... fine...

except when x is 0, 1 or infinity.

ecksor

As for the first post, you haven't actually proven anything. Can you convince us why we should accept the equality of powers like that?

Victor

You are effectively dividing by zero when you are "equating the powers" 1^1 also = 1^2, so is 1=2?

Excel says 1^0 = 0.

Sev

I was just showing how unfounded the original proof was with a much simpler but equally flawed example.

ecksor

If we're going to post flawed examples to illuminate the flaw in the original logic, then by a similar lack of reasoning I might say that

a + b = c + d = 10

a = 4
b = 6

=> c = 4
d = 6

The point being that (usually) a solution can be reached in more ways than one.

ozt9vdujny3srf

victor anything to the power of 0 is 1

Giblet

5^2/5^2 =1, now in indices when you divide something, it is the same as having the bottom indice brought to the top and changed sign. and the 5's are common, so its 5^(2-2) = 5^0 = 1

ozt9vdujny3srf

and if x^y = x^z the y = z

otherwise they would not be equal

ozt9vdujny3srf

Giblets one is better , and is true for any value of the 5 if ye no what i mean :P

Bard

Originally posted by Truckle
and if x^y = x^z the y = z

otherwise they would not be equal

like I said, this is not necessarily the case for 0, +/- 1 or +/- infinity.

Giblet

Because its fact :>

Giblet

Powers can not be used as an example

1 not multiplied by itself = 1 (1^0)
1 multiplied by itself = 1 (1^1)

Thats what squares are, so they can't be used as an example.

Victor

Originally posted by Truckle
victor anything to the power of 0 is 1

Tell Microsoft, not me.

smiles

right, to equate powers what you are actually doing to taking the natural log of the number.

so say you have 2^3
ln(2^3) = 3 * ln(2)

so usually when you have 2^x = 2^3
then x * ln(2) = 3 * ln(2)
and the ln(2) cancels, leaving x = 3


however ln(1) = 0
so you cannot use it in one of those equations as you get both sides equal to 0 regardless of the powers.

<< Fio >>

poontang

hey horse, what about this:
0=+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1..............infinity
=1-(1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1..............infinity)
=1-0
=1
i created something from nothing, like...

Giblet

Originally posted by poontang
hey horse, what about this:
0=+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1..............infinity
=1-(1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1..............infinity)
=1-0
=1
i created something from nothing, like...

1+1-1+1-1..infinity, would be undefined.

Gordon

Christ I'm so glad I didn't pay attention in Pure Maths A-level!

ozt9vdujny3srf

:P

Hi, Mark.

As Mark wrote to Dr. Math
On 04/13/2003 at 18:47:05 (Eastern Time),
>[Question]
>1^1 = 1
>
>1^0 = 1
>
>So 1^1 = 1^0
>
> By equating the powers, we get 1 = 0
>
>
>Someone take it apart please (although I believe mine to be slightly
>tougher that the other one
>
>
>[Difficulty]
>the fact that the answer is impossible :P
>
>[Thoughts]
>

You seem to be just ASSUMING this "fact":

if a^b = a^c, then b = c, for ANY a

That is true IF you can take the base-a logarithm of both sides. But
it doesn't work for a=1, because there is no base-1 logarithm, which
would be defined this way:

y = log_1(x) if 1^y = x

But since 1^y = 1 for ALL y, this logarithm can't be defined. And in
fact, that is just what you are seeing: 1^1, 1^0, and in fact 1^y for
any y are all equal to 1, though 1, 0 and any y are NOT equal.

Remember that any step you take in a proof must be justified by some
known fact; you can't just do things without thinking!

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum
<http://mathforum.org/dr.math/


WEll done Bard, I think you were cosest to that explanation :P


cheers all (especially the people i pissed off on irc )

Lump

You seem to be just ASSUMING this "fact":

if a^b = a^c, then b = c, for ANY a

That is true IF you can take the base-a logarithm of both sides. But
it doesn't work for a=1, because there is no base-1 logarithm, which
would be defined this way:

y = log_1(x) if 1^y = x

But since 1^y = 1 for ALL y, this logarithm can't be defined. And in
fact, that is just what you are seeing: 1^1, 1^0, and in fact 1^y for
any y are all equal to 1, though 1, 0 and any y are NOT equal.

Remember that any step you take in a proof must be justified by some
known fact; you can't just do things without thinking!

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum
<http://mathforum.org/dr.math/>



John

smiles

Originally posted by Truckle
:P

That is true IF you can take the base-a logarithm of both sides. But
it doesn't work for a=1, because there is no base-1 logarithm, which
would be defined this way:

y = log_1(x) if 1^y = x

But since 1^y = 1 for ALL y, this logarithm can't be defined. And in
fact, that is just what you are seeing: 1^1, 1^0, and in fact 1^y for
any y are all equal to 1, though 1, 0 and any y are NOT equal.

[....]

WEll done Bard, I think you were cosest to that explanation :P


cheers all (especially the people i pissed off on irc )

So you still ignored me!


<< Fio >>

ozt9vdujny3srf

Jesus smiles , now tha6t i think of it :P

/me gives smiles a meddle