Is the equation 0!=1 based on flawed reasoning?

[sysprog]

 

[fresh_42]

I am interested in why some Mathematicians use the formula n!=n(n-1)! - please see my original post.

We want to abbreviate $1\cdot 2\cdot 3 \cdots n$ and choose the notation $n!$ for this product. There are 3 possible ways to do so:

• by words
• by dots, like I just did
• by recursion

The latter means basically by its precedessor, i.e. $(n-1)!$ This is commonly seen as the most elegant way since it avoids dots or phrases like 'and so on'. As explained before, this needs an anchor, a starting point, which is $0!=1$ for various reasons.

 

[sysprog]

@fresh_42, so do you suppose it would be more explanatory or just more inelegant (or something else) to write $0!=0.9^* =1$ instead of just writing $0!=1$?

 

[PeroK]

@fresh_42, so do you suppose it would be more explanatory or just more inelegant (or something else) to write $0!=0.9^* =1$ instead of just writing $0!=1$?

In general, we have a universal constant $0! = f_0$, which appears as a factor in the relevant equations. E.g. $$\binom n k = \frac{(n!)f_0}{(n-k)!k!}$$For simplicity, mathematicians generally work with the choice of $f_0 = 1$.

 

[Pete Mcg]

We want to abbreviate $1\cdot 2\cdot 3 \cdots n$ and choose the notation $n!$ for this product. There are 3 possible ways to do so:

• by words
• by dots, like I just did
• by recursion

The latter means basically by its precedessor, i.e. $(n-1)!$ This is commonly seen as the most elegant way since it avoids dots or phrases like 'and so on'. As explained before, this needs an anchor, a starting point, which is $0!=1$ for various reasons.

Thank you.

 

[Pete Mcg]

Yes. I have seen this one. It's good.

 

[Pete Mcg]

Thank you. You have been very helpful.

 

[Pete Mcg]

Yup, that is the gist of it.

If you want to understand what mathematicians call a proof https://www.people.vcu.edu/~rhammack/BookOfProof/

It is pdf made free by the author. Happy reading.

Thank you so much!It all makes sense now...

 

[Pete Mcg]

Thank you so much!It all makes sense now...

My God! I've just had a peek at Book Of Proof and it's marvellous. Can I write you into my will?

 

[Mark44]

please correct me if I'm wrong. 0!=1 is a 'definition' in the same way that 1=1 or any other self evident statement is a definition and 'You do no prove a definition in Mathematics'.

1 = 1 is not a definition. In general for all kinds of definitions, including words found in a dictionary, you can't define something in terms of itself.
The equation 1 = 1 is an example of the reflexive property of the equality relation. This property says that any number is equal to itself. Other relations, such as < or >, do not have this property. For example, $5 \nless 5$ and $2 \ngtr 2$.

 

[MidgetDwarf]

My God! I've just had a peek at Book Of Proof and it's marvellous. Can I write you into my will?

Yes .

 

[Baluncore]

Throughout this thread I see; n! = n(n-1)!
But that is sloppy since the complete recursive definition of factorial is;
n! = n(n-1)! for n≥1 and 0!=1
Everything here is a definition.
Do not attempt to use the definition as a self referential proof of itself.

 

[PeroK]

$0! = 1$ (by definition)

$n! = n(n-1)!$ (for $n \ge 1$)

You obviously failed to see post #2.

 

[Baluncore]

@PeroK
I did see post #2. You got it right, but most of the thread ignores the complete definition and so becomes incomplete or self-referential.

 

[Frabjous]

@PeroK
I did see post #2. You got it right, but most of the thread ignores the complete definition and so becomes incomplete or self-referential.

I apologize for not meeting your standards.

 

[Baluncore]

I apologize for not meeting your standards.

No apology is necessary in hindsight if it gets people to think. My standards are lower than yours.

This interesting thread gave me the uncanny feeling that we were contemplating the proof of a definition, from inside a fishbowl, while looking at our different reflections in the curved wall.

Meanwhile some find the difference between antimony and an antinomy.

 

[sysprog]

I apologize for not meeting your standards.

Hmm $\dots -$ here's an observation not especially related to meeting standards, but at least ancillarily related to this thread topic: computationally, for e.g. card-deck-sized non-zero factorials, e.g. between euchre, poker, and pinochle or canasta sized decks, an iterative method is a little faster than recursive, but for zero, there's no comparison necessary between those methods, because we just hardcode zero factorial to 1 $-$ it's so by definition, as @PeroK said, and reasonably so, as @fresh_42, @PeroK, @Baluncore, @Mark44, and maybe others, explained.

 

[Baluncore]

@sysprog Is the definition of n! iterative or recursive ?
I think you are saying that, even if I knew the difference between an iterative and a recursive definition, I could not apply either to the definition of 0!

 

[pbuk]

Is the definition of n! iterative or recursive ?

Iterative (or explicit) definition:
$$ n! = \begin{cases}
\prod\limits_{k=1}^n k, & \text{if}\ n>0 \\
1, & \text{if}\ n = 0
\end{cases}
$$
Recursive (or implicit) definition:
$$ n! = \begin{cases}
n(n-1)! , & \text{if}\ n>0 \\
1, & \text{if}\ n = 0
\end{cases}
$$
So $0!$ is defined identically in each case.

 

[jbriggs444]

Or simply $$n!=\prod\limits_{k=1}^n k$$ with the understanding that a product indexed from one to zero is the empty product which is, by definition, the neutral element for multiplication.

The meaning of the empty product was already belabored up-thread.

Edit:

In the world of computing, the Ada programming language makes explicit the notion of an empty array slice such as a[1:0]. If the initial index is one greater than the final index, the result is explicitly the empty array and no bounds checking on either index is performed.

In VAX Fortran-77 the same would work and was needed to create an empty string, but one needed to compile with bounds checking disabled.

 

[Pete Mcg]

Yes .

How does 10% of almost nothing sound?

 

[Pete Mcg]

1 = 1 is not a definition. In general for all kinds of definitions, including words found in a dictionary, you can't define something in terms of itself.
The equation 1 = 1 is an example of the reflexive property of the equality relation. This property says that any number is equal to itself. Other relations, such as < or >, do not have this property. For example, $5 \nless 5$ and $2 \ngtr 2$.

Thank you for that. As stated previously in my original post, I'm not a Mathematician and am learning, every little bit counts and I appreciate your and everybody's input . (Learned today that you when you multiply a negative number by a negative number the result is a positive number - they didn't teach this in Maths @ school I went to so it took a bit of getting my head round it - at first I went HUH?? but it's starting to makes sense.) This a new world to me, good fun -talk about fun - this Godel incompleteness thing has got me intrigued!
As in learning any new 'language' one needs to understand the rules and conventions (another thing learned today was BOMDAS / PEMDAS)Even the supposedly simplest thing like 'number' has a myriad of different 'meanings' - natural numbers, rational numbers, irrational etc etc etc ...Anyway, enough of my rambling. Cheers.

 

[fresh_42]

they didn't teach this in Maths @ school I went to so it took a bit of getting my head round it - at first I went HUH??

A product is an area. An area is oriented. It makes a difference whether you circle it clockwise or counter-clockwise, one is noted as a positive number, the other one by a negative number. Which is which is up to you. The lines are oriented, too, up and down, left and right. Now $(+1)\cdot (+1)$ has the same orientation as $(-1)\cdot (-1),$ and $(+1)\cdot (-1)$ is of opposite orientation.

 

[jbriggs444]

Another motivation is the distributive law: $a \times (b+c) = a \times b + a \times c$

Say you have $-2 \times (1-1)$ then clearly that should evaluate as $-2 \times 0 = 0$. If we apply the distributive law, it should be equal to $( -2 \times 1 ) + ( -2 \times -1 )$. But if a negative times a negative is a negative, that formula evaluates as $-2 + -2$ for a result of $-4$ which is wrong.

So if a negative times a negative yields a negative, the distributive law breaks.

 

[Mark44]

(Learned today that you when you multiply a negative number by a negative number the result is a positive number - they didn't teach this in Maths @ school I went to so it took a bit of getting my head round it

Sorry to hear that it wasn't taught at your school. My first exposure to signed-number arithmetic was in ninth grade, back when I was 14.

 

[pbuk]

Another motivation is the distributive law: $a \times (b+c) = a \times b + a \times c$

So if a negative times a negative yields a negative, the distributive law breaks.

If a negative times a negative yields a negative there are no multiplicative inverses for negative numbers so we haven't even got a group.