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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.]
What else can I help you with?
How do you prove that the sum of a rational number and its additive inverse is zero?
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
Prove that the inverse of an invertible mapping is invertible?
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.
Prove that square root of 2 is an irrational number?
The square root of 2 is 1.141..... is an irrational number
How to prove that A being any number A multiplied by A plus 1 is always divisible by 2?
A will always be an odd number.
How can you prove that the square root of three is an irrational number?
Because 3 is a prime number and as such its square root is irrational
How do you prove that 1021 is the multiplicative inverse of 2.1?
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
How do you prove the multiplicative inverse is unique?
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'
Why does zero have no multiplicative inverse?
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".
Why does the fact 0 has no multiplicative inverse still mean R is still a field?
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.
How do I prove the additive inverse is unique?
assume its not. make two cases show that the two cases are equal
How do you prove the ideal bandpass filter theorem with an inverse Fourier transform?
this question on pic
How do you prove that the sum of a rational number and its additive inverse is zero?
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
How will you prove that multiplication and division are inverse processes?
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.
What models can you to prove that opposites combine to zero?
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.
What is the inverse of the statement below?
Answer this question… Which term best describes a proof in which you assume the opposite of what you want to prove?
Prove that the inverse of an invertible mapping is invertible?
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.
What did Avogadro's number prove?
no prove....
