If any number is mutiplied or devided by zero, is zero! Then why is 0 Factoral =1 ?

Answer 1

This is just a definition, to avoid the issue of division by zero, which generates infinity (or an error on your calculator). The statement 0! = 1 comes from the recurrent nature of the factorial:
n! = n*(n-1)!
(n-1)! = (n-1)*(n-2)!
...
1! = 1*0!
In order for the last equation to be true, you need to have 0! different from 0. Since the series of n! works with integer numbers, it makes sense to use the next fundamental mathematical constants, which is 1.

Answer 2

Actually it's not a "special case" definition as people seem to think. It comes right from the definition of factorial.

The definition of n! is the product of all integers less than or equal to n and greater than 0.

So 2! = 2*1, and 1!=1. We know this.

Now we need to understand the "empty product". Perhaps it's easier to start with the empty sum. The empty sum is 0 because for any addition problem a+b, we can put a 0 in front to make it 0+a+b without changing the value. So *every* sum can be rewritten to include the empty sum. In essence, the empty sum is the first term in every addition problem. (99.99% of the time there is no reason to think of it like this, but it's true and it helps the intuition for the next step).

Analogously, the empty product is 1. For any product ab, we can rewrite 1*ab without changing the value. *Every* product begins as the empty product 1 and then multiplies other numbers after it. If we want to evaluate 3*5, we begin with the empty product 1, then multiply by 3, then by 5, and we get 15.

Now 0! is, by definition, the product of all integers less than or equal to 0 and greater than 0. There are no such integers. So if we are asked to evaluate this quantity 0!, we begin with the empty product 1 and then multiply by... nothing, we just established that 0! has no factors. All we have is 1. And that's the answer.

It's the exact same reason that anything raised to the zero power is 1. If we're asked to evaluate 2^0, we start with 1 and multiply it by 2 zero times. Okay then, it's still 1. That's why n^0 = 1.

There, now I killed two common questions with one stone! I went a little fast because I'm short on time, but I'll check back probably tomorrow in case there is some confusion.

Answer 3

Any number multiplied by zero is zero. This may sound obvious with normal numbers, but it is important to remember this for other number fields as this is the definition of a zero.

Any number divided by zero is undefined (except in one very specific number field which is NOT the one we use in everyday calculations).

The factorial function that most people see with integers (e.g., 5! = 1*2*3*4*5 = 120) is only a small part of a much larger function called the Gamma function. From it, we can deduce the following equation:

n! = n* (n-1)!

To find the factorial of n, one takes the factorial of (n-1) and multiplies it by n.

If n = 1, then we have:

1! = 1 * 0! (factorial of 1 equals 1 times factorial of zero). The only way this makes sense (and the only way the Gamma function can make sense) is if 0! = 1

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using the same equation, we can have:
(n-1)! = n!/n

0! = 1!/1 = 1/1 = 1

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The Gamma function works for fractions, it works for complex numbers (with square roots of negative numbers), and so on, yet, it does not work with negative integers. So, even though we can pretend to find the factorial of 1.5 (around 1.94...), we can't find a factorial of -2.

Answer 4

0! = 1 is a definition, not a derived result.

In general for x not equal to 0, 0/x = 0, x/0 is undefined. If x = 0 then 0/0 may be 0, undefined or some finite value, depending on the context. This type of problem generally occurs when calculating a limit of some function.

Math Rules!

Answer 5

You can't divide a number by 0. 0! intuitively should represent the number of ways you can order a set with 0 items, which is 1 way.

Answer 6

As for the first statement, anything divided by zero is undefined. As for the question, I am unsure if 0! =1. However, if that is correct it is simply because it is defined that way.

Answer 7

because 1 can be divided into an infinite amount of values, in other words, drop a ball, and there are infinite fractions of seconds in time (some so close to zero it would make you feel faint) before it bounces back up again.

Answer 8

actually it had to be defined that way
0!=1 I believe there may be a proof for it, you can try googling it, but I believe at first they just had to make 0!=1 by definition.

Answer 9

we know nPn is 1. again nPn =n!/n!(n-n)! that is n!/n!*0!
equating both sides we get 0!=1 however 0! has no physical significance.

Answer 10

You cannot divide by Zero.
You can divide zero by other numbers though.