Why 1 = 0!

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.

mistervengeance

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

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

And here I was expecting to see a proof for 1 = 0, not 1 = 0!

GabuEx

1 cant = 0 and doesnt what you prooved was 1=0! which is different, 1=0 is a math error and cannot happen(for example: You cant have 1 apple and have no apples, if you have 1 apple you have 1 apple) you were close and I admire the effort but 1=/=0 but 1 can equal (0!). sparkypants

[QUOTE="sparkypants"]1 cant = 0 and doesnt what you prooved was 1=0! which is different, 1=0 is a math error and cannot happen(for example: You cant have 1 apple and have no apples, if you have 1 apple you have 1 apple) you were close and I admire the effort but 1=/=0 but 1 can equal (0!). Funky_Llama

sorry wasnt really paying attention to the forum just the 1st post and felt like showing off lol :P

M'kay...

Forgot all this math after geometry.

Think it logically if you have one dollar you have one dollar not zero. If you have one apple you have 1 apple not zero. /thread. Omni-Wrath

[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.

I was just going to say

2a=a would mean

a^2=2 and the square root of 2 is not 1

[QUOTE="loco145"][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.

sparkypants

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

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

I was just going to say

2a=a would mean

a^2=2 and the square root of 2 is not 1

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

loco145

[QUOTE="loco145"]

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

Funky_Llama

Try and prove the Riemann Hypothesis.CBR600-RR

To all of those of you who lie awake at night wondering about the factorial function...

1 = 0!

Why?

Because of the following:

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

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

which, through dividing both sides of the equation by (n+1) becomes n! = (n+1)!/(n+1)

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

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

and, since 1! is equal to 1, we get

0! = 1/1 = 1

NOTE: the factorial function n! is primarily defined as the multiplication of all numbers from 1 to n.

thepwninator

[QUOTE="thepwninator"]

To all of those of you who lie awake at night wondering about the factorial function...

1 = 0!

Why?

Because of the following:

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

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

which, through dividing both sides of the equation by (n+1) becomes n! = (n+1)!/(n+1)

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

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

and, since 1! is equal to 1, we get

0! = 1/1 = 1

NOTE: the factorial function n! is primarily defined as the multiplication of all numbers from 1 to n.

deepdreamer256

How so?

You doubt the fact that zero factorial is equal to one? Type it into your calculator and see!

Are you saying that nothing is something?

Are you saying that nothing is something?

BranKetra

Are you saying that nothing is something?

BranKetra

No. I'm proving that something that most people assume would be nothing is, in fact, something :)

[QUOTE="BranKetra"]

Are you saying that nothing is something?

loco145

[QUOTE="BranKetra"]

Are you saying that nothing is something?

thepwninator

No. I'm proving that something that most people assume would be nothing is, in fact, something :)

To all of those of you who lie awake at night wondering about the factorial function...

1 = 0!

Why?

Because of the following:

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

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

which, through dividing both sides of the equation by (n+1) becomes n! = (n+1)!/(n+1)

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

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

and, since 1! is equal to 1, we get

0! = 1/1 = 1

NOTE: the factorial function n! is primarily defined as the multiplication of all numbers from 1 to n.

thepwninator

0! = 1 because the factorial function was defined for the purpose of computing permutations and combinations. 0! = 0 would make formulas counting such things obsolete.

[QUOTE="deepdreamer256"][QUOTE="thepwninator"]

To all of those of you who lie awake at night wondering about the factorial function...

1 = 0!

Why?

Because of the following:

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

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

which, through dividing both sides of the equation by (n+1) becomes n! = (n+1)!/(n+1)

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

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

and, since 1! is equal to 1, we get

0! = 1/1 = 1

NOTE: the factorial function n! is primarily defined as the multiplication of all numbers from 1 to n.

thepwninator

How so?

You doubt the fact that zero factorial is equal to one? Type it into your calculator and see!

[QUOTE="thepwninator"][QUOTE="deepdreamer256"][QUOTE="thepwninator"]

To all of those of you who lie awake at night wondering about the factorial function...

1 = 0!

Why?

Because of the following:

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

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

which, through dividing both sides of the equation by (n+1) becomes n! = (n+1)!/(n+1)

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

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

and, since 1! is equal to 1, we get

0! = 1/1 = 1

NOTE: the factorial function n! is primarily defined as the multiplication of all numbers from 1 to n.

deepdreamer256

How so?

You doubt the fact that zero factorial is equal to one? Type it into your calculator and see!

There was nothing wrong with his algebra.

[QUOTE="BranKetra"]

Are you saying that nothing is something?

thepwninator

No. I'm proving that something that most people assume would be nothing is, in fact, something :)

There was nothing wrong with his algebra.

nbtrap1212

[QUOTE="nbtrap1212"]

There was nothing wrong with his algebra.

loco145

Wrong. He assumed that x! is correctly defined for all positive integers and correctly deduced that it implies 0!=1.

Again, what he showed is not the reason 0!=1, but merely an implication of the definition of the factorial function on a positive integer. But it is true that 0!=1.

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

There was nothing wrong with his algebra.

nbtrap1212

Wrong. He assumed that x! is correctly defined for all positive integers and correctly deduced that it implies 0!=1.

Again, what he showed is not the reason 0!=1, but merely an implication of the definition of the factorial function on a positive integer. But it is true that 0!=1.

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)! ?

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

There was nothing wrong with his algebra.

loco145

Wrong. He assumed that x! is correctly defined for all positive integers and correctly deduced that it implies 0!=1.

Again, what he showed is not the reason 0!=1, but merely an implication of the definition of the factorial function on a positive integer. But it is true that 0!=1.

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.

I think I know what you're getting at.

If "n!*(n+1)=(n+1)! is true for all non-negative integers" is his assumption, then his algebra is correct.

However, n!*(n+1)=(n+1)! is only deducible from the definition of the factorial function on all positive integers (not 0) for n being a positive integer.

In other words, it is impossible to show that 0!=1 from the definition of the factorial function on the set of positive integers alone; one must assume that n!(n+1)=(n+1)! is true for all non-negative integers (which would imply that 0! is already defined).

Like I said before, 0!=1 because that is how we have defined it.

>_> Here's the entire overarching flaw which the entire of this concept. Just because the 0 factorial is equal to 1 does not mean that 0 as a whole is equal to one, neither in Mathematics or any other field.deepdreamer256

Where did he try to prove that 0 = 1?

>_> Besides, some of your algebra is completely botched. For example: How does this (n-1)! * n =n! convert into this n! * (n+1) = (n+1)! ??? You haven't balanced both sides at all.deepdreamer256

Both formulas are correct. If you substitute n = 6 into the first one and n = 5 into the second one, you see that they're identical.

If you want to make (n+1)! the subject of the formula (which is what you have attempted to do) the only way to do so is like this: (n-1)! = n!/n Which of course, once we substitute values means 0! = 1!/1 which means 0! = 1. There's an argument much simpler than the BS you came up with.deepdreamer256

That's also valid, but it doesn't make what he did invalid.

Like I said before, 0!=1 because that is how we have defined it. nbtrap1212

Yes, but this is a very easy way to show why it was defined that way, which is an oft-asked question.

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

There was nothing wrong with his algebra.

nbtrap1212

Wrong. He assumed that x! is correctly defined for all positive integers and correctly deduced that it implies 0!=1.

Again, what he showed is not the reason 0!=1, but merely an implication of the definition of the factorial function on a positive integer. But it is true that 0!=1.

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.

I think I know what you're getting at.

If "n!*(n+1)=(n+1)! is true for all non-negative integers" is his assumption, then his algebra is correct.

However, n!*(n+1)=(n+1)! is only deducible from the definition of the factorial function on all positive integers (not 0) for n being a positive integer.

In other words, it is impossible to show that 0!=1 from the definition of the factorial function on the set of positive integers alone; one must assume that n!(n+1)=(n+1)! is true for all non-negative integers.

Like I said before, 0!=1 because that is how we have defined it.

[QUOTE="nbtrap1212"]Like I said before, 0!=1 because that is how we have defined it. GabuEx

Yes, but this is a very easy way to show why it was defined that way, which is an oft-asked question.

You're wrong. Read my last post.

0!=1 because the definition allows us to count permutations and combinations with a simple formula involving factorials. TC wasn't entirely clear on what his assumption was.

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

There was nothing wrong with his algebra.

loco145

Wrong. He assumed that x! is correctly defined for all positive integers and correctly deduced that it implies 0!=1.

Again, what he showed is not the reason 0!=1, but merely an implication of the definition of the factorial function on a positive integer. But it is true that 0!=1.

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.

I think I know what you're getting at.

If "n!*(n+1)=(n+1)! is true for all non-negative integers" is his assumption, then his algebra is correct.

However, n!*(n+1)=(n+1)! is only deducible from the definition of the factorial function on all positive integers (not 0) for n being a positive integer.

In other words, it is impossible to show that 0!=1 from the definition of the factorial function on the set of positive integers alone; one must assume that n!(n+1)=(n+1)! is true for all non-negative integers.

Like I said before, 0!=1 because that is how we have defined it.

I wouldn't agree with "it wouldn't work otherwise." But I do agree with your second sentence. Still, his algebra was correct.

That's also valid, but it doesn't make what he did invalid.GabuEx

Still, his algebra was correct.

nbtrap1212

[QUOTE="nbtrap1212"]

Still, his algebra was correct.

loco145

Sort of. The veracity of his assumption doesn't nullify his algebra, only the veracity of his conclusion.

0.999999....=1 proof. gabu ex mentioned it, and I was compelled to put it up there for ya.

x= 0.9999999.....

10x= 9.999999.....

10x-x = 9.99999... - 0.99999...

9x=9

x=1

1=0.99999999....

Dunno if this as popped up anywhere else, but there ya go.

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

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 :)

[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.

[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.nbtrap1212

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.

[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.