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.]
To prove that the residue classes modulo a prime ( p ) form a multiplicative group, consider the set of non-zero integers modulo ( p ), denoted as ( \mathbb{Z}_p^* = { 1, 2, \ldots, p-1 } ). This set is closed under multiplication since the product of any two non-zero residues modulo ( p ) is also a non-zero residue modulo ( p ). The identity element is ( 1 ), and every element ( a ) in ( \mathbb{Z}_p^* ) has a multiplicative inverse ( b ) such that ( a \cdot b \equiv 1 \mod p ) (which exists due to ( p ) being prime). Thus, ( \mathbb{Z}_p^* ) satisfies the group properties of closure, associativity, identity, and inverses, confirming it is a multiplicative group.
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.
A finite integral domain ( D ) has no zero divisors and is commutative. Since ( D ) is finite, for any non-zero element ( a \in D ), the set ( { a, 2a, 3a, \ldots, na } ) (where ( n ) is the number of elements in ( D )) must eventually repeat due to the pigeonhole principle. Thus, there exists an integer ( k ) such that ( ka = 0 ), but since ( D ) has no zero divisors, this implies ( k = 0 ) or ( a = 0 ), meaning every non-zero element has a multiplicative inverse. Therefore, ( D ) is a field.
The square root of 2 is 1.141..... is an irrational number
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
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'
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".
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.
assume its not. make two cases show that the two cases are equal
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....