Math, asked by ramyashree591, 7 months ago

What is the value of (-1)

Answers

Answered by medasanju
0

Answer:

the value remains same but it's a negative number

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Answered by Tombiju14
1

Answer:

n! is defined for any positive integer n as the product of every positive integer less than or equal to n.

Equivalently, it can be defined recursively as follows:

1! = 1.

n! = (n-1)!×n for any integer n ≥ 2.

Let us extend the recursive definition so that the rule “n! = (n-1)!×n” is true for any integer n.

Then for the case n = 1:

1! = (1–1)!×1.

So 1 = 0!×1.

So 0! = 1.

And for the case n = 0:

0! = (0–1)!×0.

So 1 = (-1)!×0.

Let d = (-1)!

Then 1 = d×0.

But given any number d, d×0 = 0. So there is no number d which satisfies the equation d×0 = 0. So d does not exist, and therefore (-1)! has no value.

So (-1)! is undefined, just as 1/0 is undefined, and for precisely the same reason.

Step-by-step explanation:

hope it helps you .

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