Why does 0! = 1 ?

[brucejin]

3! = 6
2! = 2
1! = 1
It looks like 0! = 0

Why does 0! = 1 ?

 

[Deleted member 4993]

3! = 6
2! = 2
1! = 1
It looks like 0! = 0

Why does 0! = 1 ?

This is by definition. In order to maintain definition of Combination ([sub:13sy7x4f]n[/sub:13sy7x4f]C[sub:13sy7x4f]n[/sub:13sy7x4f]) and Permutation ([sub:13sy7x4f]n[/sub:13sy7x4f]P[sub:13sy7x4f]n[/sub:13sy7x4f]) - we define 0! = 1

 

[mmm4444bot]

Subhotosh is right, on two fronts.

(1) Mathematics often requires arbitrary definitions, so that general properties and rules hold for all possible values. (Another example: defining i = sqrt[-1], so that all quadratic equations have solutions.)

(2) We don't understand why an absence of factors equals 1; we just get used to it.

Using the definition of factorials, one can at least show consistency of 0! = 1.

n! = n * (n - 1)!

This is true, for all natural numbers n.

Divide both sides by n.

n!/n = (n/n) * (n - 1)!

n!/n = (n - 1)!

This is also true, for all natural numbers n.

So, let n = 1.

1!/1 = (1 - 1)!

1 = 0!

 

[brucejin]

Thanks!

Now I have a^0 = 1 and 0! = 1.

I wonder if they are the only ones that are a little weired (at least by looking at them).

 

[brucejin]

Khan mentioned: "definition of Combination (nCn) and Permutation (nPn) ".

Where are these definitions taught? I cannot find them in middle schoold algebra text books (6 - 8th grade in USA).
Are the taught in high school?

Thanks.

 

[Deleted member 4993]

\(\displaystyle _nP_n \, = \, n! \, = \, \frac{n!}{(n-n)!}\)

\(\displaystyle _nC_n \, = \, 1 \, = \, \frac{n!}{n!\cdot (n-n)!}\)

 

[mmm4444bot]

Now I have a^0 = 1 and 0! = 1.

Be careful when writing a^0 = 1 as a definition because this equation is not true for all Real numbers a.

I continue to see or hear statements similar to "Any number raised to zero equals 1" made by educators and authors. (Emphasis not mine!) Such statements are sloppy.

What happens when a = 0 :idea:

??

I wonder if they are the only ones that are a little [weird].

I'm not sure how to quantify such weirdness, but the following is definitely weird.

We define the square root of -1 as i (which is certainly weird, in and of itself). Now multiply i times Pi. Raising Euler's constant (e) to this product yields -1.

e^(? * i) = -1

Despite the weirdness, this relationship leads to a "beautiful" equation:

e^(? * i) + 1 = 0

Some people argue that this equation relates the five most important numbers in the universe!

??

Where are [combination and permutation] definitions taught? I cannot find them in middle schoold algebra text books …

Any decent precalculus textbook covers this material; although, I wouldn't be suprised to see these concepts introduced at the middle-school level.

Cheers ~ Mark

 

[DrMike]

3! = 6
2! = 2
1! = 1
It looks like 0! = 0

Why does 0! = 1 ?

This is by definition. In order to maintain definition of Combination ([sub:15x97kdq]n[/sub:15x97kdq]C[sub:15x97kdq]n[/sub:15x97kdq]) and Permutation ([sub:15x97kdq]n[/sub:15x97kdq]P[sub:15x97kdq]n[/sub:15x97kdq]) - we define 0! = 1

?? I thought it was in order to maintain the rule that n! = n x (n-1)!

 

[soroban]

. . . \(\displaystyle \begin{array}{c}\text{Man has wondered} \\ \text{since time immemorial} \\ \text{why 1 is the value} \\ \text{of 0!} \end{array}\)

 

[Deleted member 4993]

3! = 6
2! = 2
1! = 1
It looks like 0! = 0

Why does 0! = 1 ?

This is by definition. In order to maintain definition of Combination ([sub:3s6i4t7t]n[/sub:3s6i4t7t]C[sub:3s6i4t7t]n[/sub:3s6i4t7t]) and Permutation ([sub:3s6i4t7t]n[/sub:3s6i4t7t]P[sub:3s6i4t7t]n[/sub:3s6i4t7t]) - we define 0! = 1

?? I thought it was in order to maintain the rule that n! = n x (n-1)!

Sounds perfectly logical to me.... but in my high school (pre 1965) that (meaning dealing with C & P) was the reason (excuse) given.