Suppose p and q are inverses of a number x. where x is non-zero.
Then, by definition, xp = 1 = xq
therefore xp - xq = 0
and, by the distributive property of multiplication over subtraction,
x*(p - q) = 0
Then, since x is non-zero, (p - q) = 0.
That is, p = q.
[If x = 0 then it does not have a multiplicative inverse.]
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To prove the uniqueness of the multiplicative inverse of a real number, let's assume that there are two different multiplicative inverses, say a and b, for a given real number x. This means that a * x = b * x = 1. By multiplying both sides of the equations by the common factor x, we get a = b = 1/x, which proves that the multiplicative inverse is indeed unique.
I can give you an example and prove it: eg. take the rational no. 2......hence its additive inverse ie. its opposite no. will be -2 now lets add: =(2)+(-2) =2-2 =0 it means that the opposite no.s. get cancelled and give the answer 0 this is the same case for sum of a rational no. and its opposite no. to be ZERO
Assume that f:S->T is invertible with inverse g:T->S, then by definition of invertible mappings f*g=i(S) and g*f=i(T), which defines f as the inverse of g. So g is invertible.
The square root of 2 is 1.141..... is an irrational number
A will always be an odd number.
Because 3 is a prime number and as such its square root is irrational