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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.

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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.]

Q: How do you prove the multiplicative inverse of a real number is unique?

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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

Because 3 is a prime number and as such its square root is irrational

A will always be an odd number.

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A multiplicative inverse (or reciprocal) is a number 1/x which when multiplied by x yields the multiplicative identity (1). 2.1 = 2 and 1/10 = 21/10 21/10 x 10/21 = 210/210 = 1

The multiplicative inverse is defined as: For every number a ≠ 0 there is a number, denoted by a⁻¹ such that a . a⁻¹ = a⁻¹ . a = 1 First we need to prove that any number times zero is zero: Theorem: For any number a the value of a . 0 = 0 Proof: Consider any number a, then: a . 0 + a . 0 = a . (0 + 0) {distributive law) = a . 0 {existence of additive identity} (a . 0 + a . 0) + (-a . 0) = (a . 0) + (-a . 0) = 0 {existence of additive inverse} a . 0 + (a . 0 + (-a . 0)) = 0 {Associative law for addition} a . 0 + 0 = 0 {existence of additive inverse} a . 0 = 0 {existence of additive identity} QED Thus any number times 0 is 0. Proof of no multiplicative inverse of 0: Suppose that a multiplicative inverse of 0, denoted by 0⁻¹, exists. Then 0 . 0⁻¹ = 0⁻¹ . 0 = 1 But we have just proved that any number times 0 is 0; thus: 0⁻¹ . 0 = 0 Contradiction as 0 ≠ 1 Therefore our original assumption that there exists a multiplicative inverse of 0 must be false. Thus there is no multiplicative inverse of 0. ---------------------------------------------------- That's the mathematical proof. Logically, the multiplicative inverse undoes multiplication - it is the value to multiply a result by to get back to the original number. eg 2 × 3 = 6, so the multiplicative inverse is to multiply by 1/3 so that 6 × 1/3 = 2. Now consider 2 × 0 = 0, and 3 × 0 = 0 There is more than one number which when multiplied by 0 gives the result of 0. How can the multiplicative inverse of multiplying by 0 get back to the original number when 0 is multiplied by it? In the example, it needs to be able to give both 2 and 3, and not only that, distinguish which 0 was formed from which, even though 0 is a single "number".

You can multiply each side of the multiplicative inverse equation by the other inverse to show that any two multiplicative inverses are equal. Here it is more formally. Theorem: For all x in R, there exists y in R s.t. x * y = 1. If there is a y' in R such that x * y' = 1, then y = y'. Proof: - Start with x * y = 1. - y * x = 1 (commutative) - (y * x) * y' = 1 * y' = y' - y * (x * y') = y' (associative) - y * 1 = y' (because x*y' = 1) - y = y'

assume its not. make two cases show that the two cases are equal

In the definition of a field it is only required of the non-zero numbers to have a multiplicative inverse. If we want 0 to have a multiplicative inverse, and still keep the other axioms we see (for example by the easy to prove result that a*0 = 0 for all a) that 0 = 1, now if that does not contradict the axioms defining a field (some definitions allows 0 = 1), then we still get for any number x in the field that x = 1*x = 0*x = 0, so we would get a very boring field consisting of only one element.

this question on pic

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

Mulltiplication and division are inverse processes for the same numbers involved in the operation. If my answer is not correct wait please for the edition of this question by an expert. Thank you.

There is not much to prove there; opposite numbers, by which I take you mean "additive inverse", are defined so that their sum equals zero.

Answer this question… Which term best describes a proof in which you assume the opposite of what you want to prove?

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.

no prove....