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Reply 40
Original post by ComputerMaths97
You can't arrange 0 objects, because they're not there. It's like asking "how many different ways can you eat a burger that doesn't exist" and blatantly, you can't eat a burger that doesn't exist. Therefore your reasoning is saying that there is in fact 1 way of eating a non-existent burger, by not eating it. How stupid can you be xD.


I'm cringing at your stupidity.

The way you're meant to interpret "you can arrange 0 object in 1 way" is that there is only one bijective map from the empty set back into itself. If you had more than two brain cells to rub together you'd understand.
Original post by ComputerMaths97
I know it's a definition, I'm challenging the definition you tards. Can't just say "it is because it is" when there's no reasonable proof of it being true. "Proven" scientific theories have been disproven many times, and you seem unable to accept the fact that maybe something everyone believes can be wrong - despite it happening tons of times in the past. You can't arrange 0 objects, AND 0 X anything is supposed to be 0,unless that's wrong too. Therefore 0! can only logically be 0,yet because you've been simply told otherwise, you refuse to consider opposing views. This is why humans innovate so slowly, we all just believe what we're told.


Really mature to resort to insults to cover up your ignorance. Do you know what a definition is?...Are you even at university? Please don't keep embarrassing yourself. Don't try to make yourself sound more intellectual by bringing in scientific theories, which were built on hypotheses.


You can keep thinking that 0! is 1 but you're the only person in the world who thinks so and when you look back you will laugh at your ignorance.

And what don't you get about the fact that the only arrangement of nothing is NOTHING - that gives one.

Ah, you want to apply to Cambridge, Imperial, UCL etc. I'm at one of those unis and boy will you be laughed at if you try to challenge 0! =1.

Actually, someone in my brother's class FM class said that 0! = 1 and instantly got corrected by his teacher (funnily, PhD and undergrad from Oxford, CompSci).
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(edited 8 years ago)
Original post by Zacken
I'm cringing at your stupidity.

The way you're meant to interpret "you can arrange 0 object in 1 way" is that there is only one bijective map from the empty set back into itself. If you had more than two brain cells to rub together you'd understand.


Thank you.

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Reply 43


Some people. :shakecane:
Original post by ComputerMaths97
Explain 0! = 1 or -1 x 0 = infinity.... They make no sense, they're only "true" because they have to be or the other maths is wrong


0!=1 is an example of a vacuous identity.

For example, n! is usually defined either as:
(1) 0!=1 and (n+1)!=(n+1)n!

OR

(2) n! = #{linear orders on a set having n elements}.

In case (1), the definition says 0!=1.

In case (2), 0! is the number of linear orders on a set having 0 elements. Since any linear order is a relation, we may take the empty set as our relation, and it satisfies the linear order axioms. It satisfies the axioms since they use only universal quantifiers. See: https://en.wikipedia.org/wiki/Vacuous_truth

Linear order axioms:
(1) xRx for all x in L
(2) (xRy and yRx) implies (x=y) for all x, y in L
(3) (xRy and yRz) implies (xRz) for all x, y, z in L
(4) xRy or yRX for all x, y in L.

This is the one and only subset of a set having 0 elements, so 0!=1.
(edited 8 years ago)
Reply 45
Original post by tanyapotter
Is there a mathematical proof? When I asked my teacher he just said "because it is".

Doesn't the fact that it's not there mean that there can be no way of arranging it? I don't understand how 1! and 0! can both equal 1.


Original post by ComputerMaths97
Explain 0! = 1 or -1 x 0 = infinity.... They make no sense, they're only "true" because they have to be or the other maths is wrong


Original post by TorpidPhil
They're not true.


n!=1×2×3×(n1)×n[br][br]n!n=1×2×3×(n1)[br][br]n!n=(n1)![br][br]n=10!=1!1=1 n! = 1 \times 2 \times 3 \times \cdots (n-1) \times n [br][br]\Leftrightarrow \frac{n!}{n} = 1 \times 2 \times 3 \times \cdots (n-1) [br][br]\Leftrightarrow \frac{n!}{n} = (n-1)![br][br]n = 1 \Rightarrow 0! = \frac{1!}{1} = 1
Original post by Zacken
1×0=0-1 \times 0 = 0, not infinity - you utter idiot.

Secondly, We have that Γ(t)=lima0axt1exdx\displaystyle \Gamma(t) = \lim_{a \to \infty} \int_0^{a} x^{t-1}e^{-x} \, \mathrm{d}x and n!=Γ(n+1)n! = \Gamma(n+1).

Hence, 0!=Γ(1)=lima0aexdx=1\displaystyle 0! = \Gamma(1) = \lim_{a \to \infty} \int_0^a e^{-x} \, \mathrm{d}x = 1.

Another reason is that there is only one way to re-arrange the empty set.


n! is not defined as Gamma(n+1), it just happens to equal it, given that 0! =1

So what you have shown isn't a proof since we utilize 0! = 1 to prove the gamma relationship.
A quote from my physics teacher "Because it works".
[video="youtube;Mfk_L4Nx2ZI"]https://www.youtube.com/watch?v=Mfk_L4Nx2ZI[/video]
Reply 49
Original post by Doctor_Einstein
n! is not defined as Gamma(n+1), it just happens to equal it, given that 0! =1

So what you have shown isn't a proof since we utilize 0! = 1 to prove the gamma relationship.


I wasn't quite saying it was a proof - it's not, a proof doesn't exist. It's a definition.

However, the gamma function is an extension of the factorial function, but I get where you're coming from.
Original post by ubisoft
Being not there is a way that they are arranged


You are begging the question. There is no "they" in an empty set. To arrange x, x needs to be an object that exists. You can't arrange something that does not exist.
Original post by yl95
No, the nothingness IS the arrangement. Don't call others stupid when you're simply incorrect.
Then why have my Cambridge educated lecturers been using the fact that 0! = 1 all the time? It is a definition. Look it up. Try it on a calculator. Have you tried either?

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But arranging nothingness is contradictory in our physical world. The very act of arranging something implies the existence of an object that is being arranged. I don't dispute the mathematical truth of !0 = 1, I merely dispute it as a physical truth.
Original post by Juichiro
But arranging nothingness is contradictory in our physical world. The very act of arranging something implies the existence of an object that is being arranged. I don't dispute the mathematical truth of !0 = 1, I merely dispute it as a physical truth.


I disagree.

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What statement do you disagree with?
Original post by tanyapotter
Is there a mathematical proof? When I asked my teacher he just said "because it is".

Doesn't the fact that it's not there mean that there can be no way of arranging it? I don't understand how 1! and 0! can both equal 1.


If you want to say the empty set doesn't "exist" to be rearranged then I'm pretty sure the integer 0 doesn't either, making the expression 0! meaningless.
Original post by Juichiro
What statement do you disagree with?


Arrangements upon an empty set are not contradictory.

It may be strange, or weird, but it is not contradictory.

Just as your bowl may contain one nut, or two nuts, or maybe zero nuts, a set can contain one element, or two elements, or maybe zero elements.

Now, we can name the nuts. Lets call them nut 1 and nut 2.

Bowl 1: {Nut 1,}
Bowl 2: {Nut 1, Nut 2}
Bowl 3; {}

Forgetting for a moment how many nuts there are in the bowls, how many ways may I count the nuts in them?

Bowl 1: I say "Nut 1" = 1
Bowl 2: I say "Nut 1, nut 2", OR I say "Nut 2, Nut 1" = 2
Bowl 3: I say "" (I say nothing).

If it were Bowl 4 having Nut 1, Nut 2 and Nut 3, I might say:
"Nut 1, Nut 2, Nut 3", OR
"Nut 2, Nut 1, Nut 3", OR
"Nut 3, Nut 2, Nut 1", OR
"Nut 2, Nut 3, Nut 1", OR
"Nut 2, Nut 1, Nut 3" OR
"Nut 1, Nut 3, Nut 2".

That's 6 in total, so we have 1, 2, 1, 6 for our 4 bowls.

Now, what's the earthly point of nattering on about nuts and bowls? Could we not abstract away the physical bowl and have instead collections, and rules for manipulating them, and abstract "nuts" into "elements"?

Yes we can. That is what mathematcians do.

In this abstraction, it makes no sense to say that the empty set is nothing, any more than it is sense to say that the bowl that contains no nuts is not a bowl.
Reply 56
Original post by ComputerMaths97
You can't arrange 0 objects, because they're not there. It's like asking "how many different ways can you eat a burger that doesn't exist" and blatantly, you can't eat a burger that doesn't exist. Therefore your reasoning is saying that there is in fact 1 way of eating a non-existent burger, by not eating it. How stupid can you be xD.


That's how it works. If you're too stubborn to accept it, there's no point in this thread

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Reply 57
Original post by Raiden10
Arrangements upon an empty set are not contradictory.

It may be strange, or weird, but it is not contradictory.

Just as your bowl may contain one nut, or two nuts, or maybe zero nuts, a set can contain one element, or two elements, or maybe zero elements.

Now, we can name the nuts. Lets call them nut 1 and nut 2.

Bowl 1: {Nut 1,}
Bowl 2: {Nut 1, Nut 2}
Bowl 3; {}

Forgetting for a moment how many nuts there are in the bowls, how many ways may I count the nuts in them?

Bowl 1: I say "Nut 1" = 1
Bowl 2: I say "Nut 1, nut 2", OR I say "Nut 2, Nut 1" = 2
Bowl 3: I say "" (I say nothing).

If it were Bowl 4 having Nut 1, Nut 2 and Nut 3, I might say:
"Nut 1, Nut 2, Nut 3", OR
"Nut 2, Nut 1, Nut 3", OR
"Nut 3, Nut 2, Nut 1", OR
"Nut 2, Nut 3, Nut 1", OR
"Nut 2, Nut 1, Nut 3" OR
"Nut 1, Nut 3, Nut 2".

That's 6 in total, so we have 1, 2, 1, 6 for our 4 bowls.

Now, what's the earthly point of nattering on about nuts and bowls? Could we not abstract away the physical bowl and have instead collections, and rules for manipulating them, and abstract "nuts" into "elements"?

Yes we can. That is what mathematcians do.

In this abstraction, it makes no sense to say that the empty set is nothing, any more than it is sense to say that the bowl that contains no nuts is not a bowl.


Probably the best way I've seen this explained.

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Original post by TheBBQ
n!=1×2×3×(n1)×n[br][br]n!n=1×2×3×(n1)[br][br]n!n=(n1)![br][br]n=10!=1!1=1 n! = 1 \times 2 \times 3 \times \cdots (n-1) \times n [br][br]\Leftrightarrow \frac{n!}{n} = 1 \times 2 \times 3 \times \cdots (n-1) [br][br]\Leftrightarrow \frac{n!}{n} = (n-1)![br][br]n = 1 \Rightarrow 0! = \frac{1!}{1} = 1


Whilst 0! = 1, that is certainly not a proof of it sorry :P

Your implications are true n0\forall n \neq 0 as you divide by n in [br]n!=1×2×3×(n1)×nn!n=1×2×3×(n1)[br]n! = 1 \times 2 \times 3 \times \cdots (n-1) \times n\Rightarrow \frac{n!}{n} = 1 \times 2 \times 3 \times \cdots (n-1) which you'll know is not allowed. It looked like a nice proof otherwise, but it's incorrect (and fatally so if you're trying to prove 0! = 1).
Reply 59
If it wasn't for maths you wouldn't have been able to use this forum, the factorial example you thought it made no sense actually is an important factor in algorithms and encryption, it plays an important role in communication security.Your smart-phone relies at some stages on factorials, also on other branches such as calculus, linear algebra, probability and statistics.factorials used for encrypting and encoding also in some hardware logic, linear algebra models used for analyzing wireless communication, consider OFDM AND MIMO technology.We combine the last with calculus to evaluate the average power and energyAlso we use in each stage through the chennel statistics to estimate the perfomance of communication system.All of this depends on complex analysis If it wasn't for mathematical modelling you wouldn't have seen companies like apple, and microsoft...etc

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