Why 1 = 0!

[QUOTE="thepwninator"][QUOTE="nbtrap1212"][QUOTE="thepwninator"]

[QUOTE="GTA_dude"]I've seen these equation years ago. So thank you for finding stuff on other web sites and posting it here as yours.loco145

I never said it was mine, now did I? My dad showed this to me a few years ago, and that's the only place I've seen it :)

I was, however, asked to show this logic on the blackboard in my math cIass yesterday by my professor, as he, being more of a theoretical mathematician, did not actually believe me :)

You're only correct if you assume n!(n+1)=(n+1)! is defined for all non-negative integers (which can't be proved unless you assume 0!=1 in the first place). So you're really not proving anything.

The difference between this and the 2=1 "proof" is that nowhere in here do I divide by zero. I simply manipulate a general equation and then plug in a value for n. If you plug in -1 rather than zero, however, you divide by zero, which is not a viable thing to do, and, since the definition used in the OP necessitates a higher term, no values of the function exist for negative numbers.

The beginning assumption was that it was defined for all positive integers, not all non-negative ones. From there, it was shown that the function has a value for n = 0.

No assumptions whatsoever for the value of 0! exist in the OP.

Yeah, I don't sit awake at night wondering this.

[QUOTE="nbtrap1212"][QUOTE="thepwninator"]

[QUOTE="GTA_dude"]I've seen these equation years ago. So thank you for finding stuff on other web sites and posting it here as yours.thepwninator

I never said it was mine, now did I? My dad showed this to me a few years ago, and that's the only place I've seen it :)

I was, however, asked to show this logic on the blackboard in my math cIass yesterday by my professor, as he, being more of a theoretical mathematician, did not actually believe me :)

You're only correct if you assume n!(n+1)=(n+1)! is defined for all non-negative integers (which can't be proved unless you assume 0!=1 in the first place). So you're really not proving anything.

The difference between this and the 2=1 "proof" is that nowhere in here do I divide by zero. I simply manipulate a general equation and then plug in a value for n. If you plug in -1 rather than zero, however, you divide by zero, which is not a viable thing to do, and, since the definition used in the OP necessitates a higher term, no values of the function exist for negative numbers.

The beginning assumption was that it was defined for all positive integers, not all non-negative ones. From there, it was shown that the function has a value for n = 0.

1. Your algebra was fine.

2. If you only assume the definition of the factorial function for positive integers, you cannot possibly show that n!*(n+1)=(n+1)! for all non-negative integers.

3. Your logical error occurs when you plug n=0 into n!(n+1)=(n+1)! You cannot do that, because the formula is only true for positive integers (its actually true for n=0 too, but that's what you're trying to prove; the truth is, its only true for n=0 because we define it that way).

No assumptions whatsoever for the value of 0! exist in the OP.

thepwninator

Than you can't do this:

"Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)"

Also, in the 2=1 proof he didn't divided by 0. Is just that the cancellation isn't true if one of the factors is zero. That is true for a Ring, where the division isn't even defined.

[QUOTE="thepwninator"]

No assumptions whatsoever for the value of 0! exist in the OP.

loco145

Than you can't do this:

"Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)"

Also, in the 2=1 proof he didn't divided by 0. Is just that the cancellation isn't true if one of the factors is zero. That is true for a Ring, where the division isn't even defined.

But you can't plug n=0 into n!(n+1)=(n+1)! because n!(n+1)=(n+1)! isn't necessarily true for n=0!

[QUOTE="thepwninator"]

No assumptions whatsoever for the value of 0! exist in the OP.

loco145

Than you can't do this:

"Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)"

Also, in the 2=1 proof he didn't divided by 0. Is just that the cancellation isn't true if one of the factors is zero. That is true for a Ring, where the division isn't even defined.

Cancellation works by dividing both sides by that which you are cancelling, and the value of that which was cancelled in that proof was zero.

And that was not an application of an assumption; that was simply the plugging in a value. If you do so for any negative value, it is completely undefined. It was simply showing that the equation works for zero as well as all positive integers, thus extending its assumed realm of validity.

But you can't plug n=0 into n!(n+1)=(n+1)! because n!(n+1)=(n+1)! isn't necessarily true for n=0!

nbtrap1212

This too :)

[QUOTE="loco145"][QUOTE="thepwninator"]

No assumptions whatsoever for the value of 0! exist in the OP.

thepwninator

Than you can't do this:

"Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)"

Also, in the 2=1 proof he didn't divided by 0. Is just that the cancellation isn't true if one of the factors is zero. That is true for a Ring, where the division isn't even defined.

Cancellation works by dividing both sides by that which you are cancelling, and the value of that which was cancelled in that proof was zero.

And that was not an application of an assumption; that was simply the plugging in a value. If you do so for any negative value, it is completely undefined. It was simply showing that the equation works for zero as well as all positive integers, thus extending its assumed realm of validity.

But you can't plug n=0 into n!(n+1)=(n+1)! because n!(n+1)=(n+1)! isn't necessarily true for n=0!

nbtrap1212

This too :)

I was talking to you, pwninator. Your logic is flawed.

[QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"]

No assumptions whatsoever for the value of 0! exist in the OP.

nbtrap1212

Than you can't do this:

"Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)"

Also, in the 2=1 proof he didn't divided by 0. Is just that the cancellation isn't true if one of the factors is zero. That is true for a Ring, where the division isn't even defined.

Cancellation works by dividing both sides by that which you are cancelling, and the value of that which was cancelled in that proof was zero.

And that was not an application of an assumption; that was simply the plugging in a value. If you do so for any negative value, it is completely undefined. It was simply showing that the equation works for zero as well as all positive integers, thus extending its assumed realm of validity.

But you can't plug n=0 into n!(n+1)=(n+1)! because n!(n+1)=(n+1)! isn't necessarily true for n=0!

nbtrap1212

This too :)

I was talking to you, pwninator. Your logic is flawed.

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

[QUOTE="nbtrap1212"][QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"]

No assumptions whatsoever for the value of 0! exist in the OP.

thepwninator

Than you can't do this:

"Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)"

Also, in the 2=1 proof he didn't divided by 0. Is just that the cancellation isn't true if one of the factors is zero. That is true for a Ring, where the division isn't even defined.

Cancellation works by dividing both sides by that which you are cancelling, and the value of that which was cancelled in that proof was zero.

And that was not an application of an assumption; that was simply the plugging in a value. If you do so for any negative value, it is completely undefined. It was simply showing that the equation works for zero as well as all positive integers, thus extending its assumed realm of validity.

But you can't plug n=0 into n!(n+1)=(n+1)! because n!(n+1)=(n+1)! isn't necessarily true for n=0!

nbtrap1212

This too :)

I was talking to you, pwninator. Your logic is flawed.

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

Cancellation works by dividing both sides by that which you are cancelling, and the value of that which was cancelled in that proof was zero.

thepwninator

[QUOTE="thepwninator"]

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

nbtrap1212

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

Statement two does indeed not imply that 0! = (0+1)!/(0+1), but, since it is a general solution, we can plug in any number we wish and see whether or not it works, and, as I said, zero does, in fact, work in this case. You can plug in -1 and it does not work, thus derailing the function for all values less than zero. However, the great thing about general solutions is that you can plug any number you want into them. Some number may not actually work, but you can still plug those number into them.

[QUOTE="nbtrap1212"][QUOTE="thepwninator"]

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

thepwninator

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

Statement two does indeed not imply that 0! = (0+1)!/(0+1), but, since it is a general solution, we can plug in any number we wish and see whether or not it works, and, as I said, zero does, in fact, work in this case. You can plug in -1 and it does not work, thus derailing the function for all values less than zero. However, the great thing about general solutions is that you can plug any number you want into them. Some number may not actually work, but you can still plug those number into them.

[QUOTE="thepwninator"][QUOTE="nbtrap1212"][QUOTE="thepwninator"]

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

loco145

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

Statement two does indeed not imply that 0! = (0+1)!/(0+1), but, since it is a general solution, we can plug in any number we wish and see whether or not it works, and, as I said, zero does, in fact, work in this case. You can plug in -1 and it does not work, thus derailing the function for all values less than zero. However, the great thing about general solutions is that you can plug any number you want into them. Some number may not actually work, but you can still plug those number into them.

Redirected from general solution:

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Redirected from general solution:

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thepwninator

[QUOTE="thepwninator"]

Redirected from general solution:

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loco145

'Twas a joke on my part, of sorts, actually.

Anyway, I was wrong in calling it a general solution. It was simply an equation that allows you to determine the factorial of a number n. As with all equations, you can plug in any number you want. The problem is that not all numbers yield real solutions. In this case, however, zero does.

[QUOTE="nbtrap1212"][QUOTE="thepwninator"]

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

thepwninator

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

Statement two does indeed not imply that 0! = (0+1)!/(0+1), but, since it is a general solution, we can plug in any number we wish and see whether or not it works, and, as I said, zero does, in fact, work in this case. You can plug in -1 and it does not work, thus derailing the function for all values less than zero. However, the great thing about general solutions is that you can plug any number you want into them. Some number may not actually work, but you can still plug those number into them.

What do you mean "it works"? It's called logic. Just because you can plug in n=0 doesn't mean what you get is true. I have to go, I'll argue with you later.

[QUOTE="loco145"][QUOTE="thepwninator"]

Redirected from general solution:

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thepwninator

'Twas a joke on my part, of sorts, actually.

Anyway, I was wrong in calling it a general solution. It was simply an equation that allows you to determine the factorial of a number n. As with all equations, you can plug in any number you want. The problem is that not all numbers yield real solutions. In this case, however, zero does.

"In this case, however, zero does."

That's because we've already defined 0! to equal 1.

[QUOTE="thepwninator"][QUOTE="nbtrap1212"][QUOTE="thepwninator"]

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

nbtrap1212

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

Statement two does indeed not imply that 0! = (0+1)!/(0+1), but, since it is a general solution, we can plug in any number we wish and see whether or not it works, and, as I said, zero does, in fact, work in this case. You can plug in -1 and it does not work, thus derailing the function for all values less than zero. However, the great thing about general solutions is that you can plug any number you want into them. Some number may not actually work, but you can still plug those number into them.

What do you mean "it works"? It's called logic. Just because you can plug in n=0 doesn't mean what you get is true. I have to go, I'll argue with you later.

Math is completely infallible; if you get an answer, it is a completely viable answer to the question. If such were not the case, math, quite simply, would not work.

[QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"]

Redirected from general solution:

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nbtrap1212

'Twas a joke on my part, of sorts, actually.

Anyway, I was wrong in calling it a general solution. It was simply an equation that allows you to determine the factorial of a number n. As with all equations, you can plug in any number you want. The problem is that not all numbers yield real solutions. In this case, however, zero does.

"In this case, however, zero does."

That's because we've already defined 0! to equal 1.

No. It is because zero yields a solution, not because we've already defined it.

Nowhere in the logic in the OP is it assumed that 0! is equal to 1. It is simply shown.

[QUOTE="nbtrap1212"][QUOTE="thepwninator"][QUOTE="nbtrap1212"][QUOTE="thepwninator"]

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

thepwninator

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

Statement two does indeed not imply that 0! = (0+1)!/(0+1), but, since it is a general solution, we can plug in any number we wish and see whether or not it works, and, as I said, zero does, in fact, work in this case. You can plug in -1 and it does not work, thus derailing the function for all values less than zero. However, the great thing about general solutions is that you can plug any number you want into them. Some number may not actually work, but you can still plug those number into them.

What do you mean "it works"? It's called logic. Just because you can plug in n=0 doesn't mean what you get is true. I have to go, I'll argue with you later.

Math is completely infallible; if you get an answer, it is a completely viable answer to the question. If such were not the case, math, quite simply, would not work.

Oh man :( I had such high expectations. I already know that 0! = 1. I thought I'd see some formulas that proved that 1=0 -_-Hungry_bunny

Sorry to disappoint ;)

Aside: I would never have guessed that this thread would yield over 100 responses :o

[QUOTE="thepwninator"][QUOTE="nbtrap1212"][QUOTE="thepwninator"][QUOTE="nbtrap1212"]

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

loco145

Statement two does indeed not imply that 0! = (0+1)!/(0+1), but, since it is a general solution, we can plug in any number we wish and see whether or not it works, and, as I said, zero does, in fact, work in this case. You can plug in -1 and it does not work, thus derailing the function for all values less than zero. However, the great thing about general solutions is that you can plug any number you want into them. Some number may not actually work, but you can still plug those number into them.

What do you mean "it works"? It's called logic. Just because you can plug in n=0 doesn't mean what you get is true. I have to go, I'll argue with you later.

Math is completely infallible; if you get an answer, it is a completely viable answer to the question. If such were not the case, math, quite simply, would not work.

Can you point out where?

[QUOTE="loco145"][QUOTE="thepwninator"][QUOTE="nbtrap1212"][QUOTE="thepwninator"][QUOTE="nbtrap1212"][QUOTE="thepwninator"]

What's ironic about that is that you proved my assertion that it was not initially assumed that 0!=1, and, thus, by trying to point out a flaw in my logic, proved me right to someone else.

And how is it not flawed initially? It is because the equation is a general solution, and, therefore, can be manipulated like a general solution. Once we get to the point where n! is isolated, we can attempt to plug in any value and see whether or not it works, and it just so happens that zero does, in fact, work.

thepwninator

You're completely wrong. Let's start from the beginning.

1. Assume n!(n+1)=(n+1)! for all positive intergers n.

2. Statement 1 implies that n!=(n+1)!/(n+1) for all positive integers n.

3. Statement 2 does not imply that 0!=(0+1)!/(0+1) since 0 is not a positive integer.

Yeah, it's that simple.

Statement two does indeed not imply that 0! = (0+1)!/(0+1), but, since it is a general solution, we can plug in any number we wish and see whether or not it works, and, as I said, zero does, in fact, work in this case. You can plug in -1 and it does not work, thus derailing the function for all values less than zero. However, the great thing about general solutions is that you can plug any number you want into them. Some number may not actually work, but you can still plug those number into them.

What do you mean "it works"? It's called logic. Just because you can plug in n=0 doesn't mean what you get is true. I have to go, I'll argue with you later.

Math is completely infallible; if you get an answer, it is a completely viable answer to the question. If such were not the case, math, quite simply, would not work.

Can you point out where?

[QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"]

Math is completely infallible; if you get an answer, it is a completely viable answer to the question. If such were not the case, math, quite simply, would not work.

loco145

Can you point out where?

No.

n! can be defined as (n-1)! * n =n!.

Ok

which can be expressed as n! * (n+1) = (n+1)!

For 0?, why?

Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)

For 0?, why?

Rest my case.

Kind like the falacy that the other guy posted. You can't use the cancelation if one of the factors is zero. How can he asure me he can do that with (0)! ?

loco145

The fallacy in the 2=1 "proof" is that he is, in fact, dividing by zero when he cancels out the (a-b)'s. Nowhere in this do we divide by zero. We divide by n+1, which is not factorialized, which is, beyond all shadow of a doubt, 1 when n = 0.

I can't help but feel as though I've already addressed this...

The fallacy in the 2=1 "proof" is that he is, in fact, dividing by zero when he cancels out the (a-b)'s. Nowhere in this do we divide by zero. We divide by n+1, which is not factorialized, which is, beyond all shadow of a doubt, 1 when n = 0.

I can't help but feel as though I've already addressed this...

thepwninator

Lets use your definition:

n!=(n-1)!*n hence 0!=(0-1)!*0 =0. That is ignoring the fact that (-1)! isn't defined, oh well.

You see, you can't even use that definition for n!. So, the rest of your algebra doens't make sense when n=0.

n! * (n+1) = (n+1)!=>n!=(n+1)!/(n+1)

ok.

0!=(1)!/1=> 0=1????

That's why n!=(n-1)!*n doesn't apply when n=0.

[QUOTE="loco145"][QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"]

Math is completely infallible; if you get an answer, it is a completely viable answer to the question. If such were not the case, math, quite simply, would not work.

thepwninator

Can you point out where?

No.

n! can be defined as (n-1)! * n =n!.

Ok

which can be expressed as n! * (n+1) = (n+1)!

For 0?, why?

Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)

For 0?, why?

Rest my case.

Kind like the falacy that the other guy posted. You can't use the cancelation if one of the factors is zero. How can he asure me he can do that with (0)! ?

loco145

The fallacy in the 2=1 "proof" is that he is, in fact, dividing by zero when he cancels out the (a-b)'s. Nowhere in this do we divide by zero. We divide by n+1, which is not factorialized, which is, beyond all shadow of a doubt, 1 when n = 0.

I can't help but feel as though I've already addressed this...

Why he divides by 0? I don't get it :(, the error I see in the 2=1 proof is (a+b)=b, then a=0, then 0+b=b, then 0=0?, I don't get what I just did :P

EDIT: oh I just got it, if a=0 then he did 0/0 :|, but can someone explain me where he does come up with a+a=a?

[QUOTE="thepwninator"]

The fallacy in the 2=1 "proof" is that he is, in fact, dividing by zero when he cancels out the (a-b)'s. Nowhere in this do we divide by zero. We divide by n+1, which is not factorialized, which is, beyond all shadow of a doubt, 1 when n = 0.

I can't help but feel as though I've already addressed this...

loco145

Lets use your definition:

n!=(n-1)!*n hence 0!=(0-1)!*0 =0. That is ignoring the fact that (-1)! isn't defined, oh well.

You see, you can't even use that definition for n!. So, the rest of your algebra doens't make sense when n=0.

While it is true that the initial definition for n! is undefined at zero, the definition can be manipulated through completely sound algebra to n! = (n+1)!/(n+1), which IS defined at zero.

[QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"]

Math is completely infallible; if you get an answer, it is a completely viable answer to the question. If such were not the case, math, quite simply, would not work.

-KinGz-

Can you point out where?

No.

n! can be defined as (n-1)! * n =n!.

Ok

which can be expressed as n! * (n+1) = (n+1)!

For 0?, why?

Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)

For 0?, why?

Rest my case.

Kind like the falacy that the other guy posted. You can't use the cancelation if one of the factors is zero. How can he asure me he can do that with (0)! ?

loco145

The fallacy in the 2=1 "proof" is that he is, in fact, dividing by zero when he cancels out the (a-b)'s. Nowhere in this do we divide by zero. We divide by n+1, which is not factorialized, which is, beyond all shadow of a doubt, 1 when n = 0.

I can't help but feel as though I've already addressed this...

Why he divides by 0? I don't get it :(, the error I see in the 2=1 proof is (a+b)=b, then a=0, then 0+b=b, then 0=b, then a=0, then 0=0 :\

NOTE: problem line is denoted with a pair of slashes

a = b

a^2 = a*b

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

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

(a+b) = b //in order to get here, you are dividing both sides by (a-b), but, if a = b, then that is equal to zero, and you can't divide by zero.

a+a = a

2a = a

2 = 1

PROOF 2=1

[QUOTE="loco145"][QUOTE="thepwninator"]

The fallacy in the 2=1 "proof" is that he is, in fact, dividing by zero when he cancels out the (a-b)'s. Nowhere in this do we divide by zero. We divide by n+1, which is not factorialized, which is, beyond all shadow of a doubt, 1 when n = 0.

I can't help but feel as though I've already addressed this...

thepwninator

Lets use your definition:

n!=(n-1)!*n hence 0!=(0-1)!*0 =0. That is ignoring the fact that (-1)! isn't defined, oh well.

You see, you can't even use that definition for n!. So, the rest of your algebra doens't make sense when n=0.

While it is true that the initial definition for n! is undefined at zero, the definition can be manipulated through completely sound algebra to n! = (n+1)!/(n+1), which IS defined at zero.

[QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"][QUOTE="loco145"][QUOTE="thepwninator"]

Math is completely infallible; if you get an answer, it is a completely viable answer to the question. If such were not the case, math, quite simply, would not work.

-KinGz-

Can you point out where?

No.

n! can be defined as (n-1)! * n =n!.

Ok

which can be expressed as n! * (n+1) = (n+1)!

For 0?, why?

Thus, when we plug in n when n = 0, we get

0! = (0+1)!/(0+1)

For 0?, why?

Rest my case.

Kind like the falacy that the other guy posted. You can't use the cancelation if one of the factors is zero. How can he asure me he can do that with (0)! ?

loco145

The fallacy in the 2=1 "proof" is that he is, in fact, dividing by zero when he cancels out the (a-b)'s. Nowhere in this do we divide by zero. We divide by n+1, which is not factorialized, which is, beyond all shadow of a doubt, 1 when n = 0.

I can't help but feel as though I've already addressed this...

Why he divides by 0? I don't get it :(, the error I see in the 2=1 proof is (a+b)=b, then a=0, then 0+b=b, then 0=b, then a=0, then 0=0 :\

NOTE: problem line is denoted with a pair of slashes

a = b

a^2 = a*b

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

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

(a+b) = b //in order to get here, you are dividing both sides by (a-b), but, if a = b, then that is equal to zero, and you can't divide by zero.

a+a = a

2a = a

2 = 1

PROOF 2=1

Oh I see.

Lets use your definition:

n!=(n-1)!*n hence 0!=(0-1)!*0 =0. That is ignoring the fact that (-1)! isn't defined, oh well.

You see, you can't even use that definition for n!. So, the rest of your algebra doens't make sense when n=0.

n! * (n+1) = (n+1)!=>n!=(n+1)!/(n+1)

ok.

0!=(1)!/1=> 0=1????

That's why n!=(n-1)!*n doesn't apply when n=0.

loco145

This was never a proof that zero was equal to one-it was a proof that zero factorial is equal to one :)

[QUOTE="mistervengeance"]

no... that is dumb.

there is a much better one.

a = b

a^2 = a*b

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

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

(a+b) = b

a+a = a

2a = a

2 = 1

PROOF 2=1

of course, not really. cookie to the first guy who spots the error.

loco145

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

That's (a+b)*0=b*0. Thus, the next doesn't follow.

YO UWIN

COOKIE

Oh man :( I had such high expectations. I already know that 0! = 1. I thought I'd see some formulas that proved that 1=0 -_-Hungry_bunny

[QUOTE="Hungry_bunny"]Oh man :( I had such high expectations. I already know that 0! = 1. I thought I'd see some formulas that proved that 1=0 -_-Jandurin

If 1 = 0, math would now work in any way whatsoever; you could set any number equal to another through a bit of algebra-ing :|

[QUOTE="Jandurin"][QUOTE="Hungry_bunny"]Oh man :( I had such high expectations. I already know that 0! = 1. I thought I'd see some formulas that proved that 1=0 -_-thepwninator

If 1 = 0, math would now work in any way whatsoever; you could set any number equal to another through a bit of algebra-ing :|

[QUOTE="thepwninator"]

[QUOTE="Jandurin"][QUOTE="Hungry_bunny"]Oh man :( I had such high expectations. I already know that 0! = 1. I thought I'd see some formulas that proved that 1=0 -_-Jandurin

If 1 = 0, math would now work in any way whatsoever; you could set any number equal to another through a bit of algebra-ing :|

Or implosion.

It would be like dividing by zero, only worse :?

While it is true that the initial definition for n! is undefined at zero, the definition can be manipulated through completely sound algebra to n! = (n+1)!/(n+1), which IS defined at zero. thepwninator

[QUOTE="thepwninator"]While it is true that the initial definition for n! is undefined at zero, the definition can be manipulated through completely sound algebra to n! = (n+1)!/(n+1), which IS defined at zero. loco145

No, I am not saying that 1 is equal to zero. I am saying that zero factorial is equal to 1; there's a difference :)

0=1.

Proof:

n!=(n-1)!*n hence 0!=(0-1)!*0 =0.

Also,

n! * (n+1) = (n+1)!=>n!=(n+1)!/(n+1)

=> 0!=(0+1)!/(0+1) =1.

Hence, 0=1.

Spot the mistake.

0=1.

Proof:

n!=(n-1)!*n hence 0!=(0-1)!*0 =0.

Also,

n! * (n+1) = (n+1)!=>n!=(n+1)!/(n+1)

=> 0!=(0+1)!/(0+1) =1.

Hence, 0=1.

Spot the mistake.

loco145

TC you're wrong. Your logic fails. You lose.

The fact is, you can prove ANYTHING if you're assumption is false. and this is a great example.

That's strange. I have another way to make one = zero but it does involve infinity which isn't really a real number.

infinity+1=infinity

infinity-infinity=0

(infinity+1)-infinity=0

1+infinity-infinity=0

1=0

Okay that's really bad algebra but you get the idea.

That's strange. I have another way to make one = zero but it does involve infinity which isn't really a real number.

infinity+1=0

infinity-infinity=0

(infinity+1)-infinity=0

1+infinity-infinity=0

1=0

Okay that's really bad algebra but you get the idea.

domatron23

I don't follow.

And the TC tried to prove 0!=1 not 0=1.

My head hurts. :(Stevo_the_gamer

Nice sig.

infinity+1=0domatron23

Is this the new math? :P

[QUOTE="domatron23"]infinity+1=0GabuEx

Is this the new math? :P

Its make believe math!

That was meant to be a minus sign, i'll fix it up.

[QUOTE="GabuEx"]

[QUOTE="domatron23"]infinity+1=0domatron23

Is this the new math? :P

Its make believe math!

That was meant to be a minus sign, i'll fix it up.

Oops I fail. I meant the zero was meant to beinfinity. It's all fixed up now but still faulty.