1 = 0

ozt9vdujny3srf · 13-04-2003 10:26PM #1

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 · 13-04-2003 10:39PM

1! = 0!
=> 1 = 0

Maths is only rules.

Lorax · 13-04-2003 10:46PM

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

my attempt anyway!

ozt9vdujny3srf · 13-04-2003 11:03PM

you can't cancel a factorial afaik

ecksor · 13-04-2003 11:11PM

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 · 13-04-2003 11:13PM

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

Yeah, sure ... fine...

except when x is 0, 1 or infinity.

ecksor · 13-04-2003 11:16PM

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 · 13-04-2003 11:16PM

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 · 13-04-2003 11:18PM

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

ecksor · 13-04-2003 11:21PM

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 · 13-04-2003 11:24PM

victor anything to the power of 0 is 1

Giblet · 13-04-2003 11:26PM

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 · 13-04-2003 11:27PM

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

otherwise they would not be equal

ozt9vdujny3srf · 13-04-2003 11:29PM

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

Bard · 13-04-2003 11:31PM

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 · 13-04-2003 11:32PM

Because its fact :>

Giblet · 13-04-2003 11:35PM

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 · 13-04-2003 11:36PM

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

Tell Microsoft, not me.

smiles · 13-04-2003 11:50PM

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 · 14-04-2003 11:53AM

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 · 14-04-2003 01:47PM

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 · 14-04-2003 01:56PM

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

ozt9vdujny3srf · 14-04-2003 05:46PM

: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 · 14-04-2003 08:50PM

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 · 14-04-2003 09:27PM

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 · 14-04-2003 09:34PM

Jesus smiles , now tha6t i think of it :P

/me gives smiles a meddle

1 = 0

Welcome!

Comments

Welcome!