Why does zero have no multiplicative inverse?

Answer 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.

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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".