No, it does not.
For every real number, x, which is not zero, there exists a real number x' such that x * x' = x' * x = 1, the multiplicative identity.
Multiplicative inverse.
For any two non zero integers a and b, a/b x b/a = 1 ------------------------------------- For every number a ≠ 0 there is a number denoted by a^-1 (a to the power -1) such that: a × a^-1 = a^-1 × a = 1
1
The multiplicative inverse is when you multiply a certain number, and the product is itself, the number. So, the multiplicative inverse of 8 is of course, 1. For every number, the multiplicative number is 1, because a certain number times 1 is equal to the certain number. It's simple!!
Every non zero number has a multiplicative inverse, which is 1 divided by that number. This stands for both real and complex numbers. This can be proved by letting x=some non zero number. x*(1/x)=x/x=1, therefore the multiplicative inverse of x is 1/x.
No, it does not.
Yes
For every real number, x, which is not zero, there exists a real number x' such that x * x' = x' * x = 1, the multiplicative identity.
Multiplicative inverse.
For any two non zero integers a and b, a/b x b/a = 1 ------------------------------------- For every number a ≠ 0 there is a number denoted by a^-1 (a to the power -1) such that: a × a^-1 = a^-1 × a = 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".
A multiplicative inverse of 5 mod7 would be a number n ( not a unique one) such that 5n=1Let's look at the possible numbers5x1=5mode 75x2=10=3 mod 75x3=15=1 mod 7 THAT WILL DO IT3 is the multiplicative inverse of 5 mod 7.What about the others? 5x4=20, that is -1 mod 7 or 65x5=25 which is 4 mod 75x6=30 which is -5 or 2 mod 7How did we know it existed? Because 7 is a prime. For every prime number p and positive integer n, there exists a finite field with pn elements. This is an important theorem in abstract algebra. Since it is a field, it must have a multiplicative inverse. So the numbers mod 7 make up a field and hence have a multiplicative inverse.
1
You can invert almost any number by dividing 1 by that number. Zero is an exception since division by zero yields the equivalent of infinity, which is difficult to deal with by the usual rules of arithmetic. We cannot really know what the product of zero and infinity is. All other real numbers can be inverted.
For every number, a,there exists a number called the additive inverse, -a, such that a + -a = 0.