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Q: Why doesn't every real number have a multiplicative inverse?

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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!!

No, it does not.

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.

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.

No, zero does not. Multiplicative inverse, also known as reciprocal, is a number which multiplied by the original number gives 1 for the answer. Zero, multiplied by any numberequals zero. Infinity is not an actual number that you can multiply by. These are important concepts.

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 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.

For every number, a,there exists a number called the additive inverse, -a, such that a + -a = 0.

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