0

We consider following equation a×0=0 in this equation we able to multiply by zero. but why not divided by zero? when multiplication and division are just inverse process.

Vinod Choudhary
  • 19
  • 2
    Multiplication by $0$ has no inverse process. You can't very well say that $\frac{a\cdot 0}{0}=a$ for all $a$. You end up with $\frac{0}{0}=a$ for all $a$. But division needs to be defined so as to have a unique answer for any given pair of inputs. – paw88789 Oct 14 '14 at 06:55
  • Division by 0 or even division by banas is possible. As you can see in your own example the reverse of what is multiplied by 0 to give 0 is every number . The problem is with the uniqueness requirement, so it is true to say that the result of dividing 0 by 0 is every number. – jimjim Jan 20 '22 at 00:02

4 Answers4

5

Division, by definition, is multiplying by an inverse element. This means that $\frac{a}{b}$ is defined to be equal to $a\cdot b^{-1}$, and $b^{-1}$ is defined as

$b^{-1}$ is the unique real number that satisfies $b\cdot b^{-1}=1$

The problem with dividing by $0$ is that $0^{-1}$ does not exist, because if $x=0^{-1}$ did exist, then it would satisfy the equation $0\cdot x = 1$, but since $0\cdot x=0$ for all $x$ and $0\neq 1$, this is impossible.

5xum
  • 123,496
  • 6
  • 128
  • 204
4

This example shows that multiplication and division are actually not inverse processes. "Partially inverse processes" would be a better description. More precisely, multiplication by zero is not an invertible operation, because we can find many inputs that all give the same result when multiplied by zero.

In fact, it's worth noting that multiplication by 2 is not invertible in the complementary sense, when we're working in the integers: e.g. $2x = 3$ has no solutions for $x$ when we are working in the integers.

The point is that multiplication can only be partially inverted: "multiplication by $m$" can only be inverted if $m$ is a "unit": that is, if we can solve the equation $mx = 1$ for $x$.

Also, multiplication by $m$ can be partially inverted if $m$ is 'cancellable': that is, if the only solution to $mx = 0$ is $x=0$. If we had two solutions $x=a$ and $x=b$ to the solution $mx=c$, then $m(a-b)=ma - mb = c - c = 0$ and therefore $a=b$. Thus in this case, solutions are unique if they exist, although they aren't guaranteed to exist. (e.g. my multiplication by $2$ in the integers example above)

2

Since $a\cdot 0=0$ and $b \cdot 0=0$, then $\frac{0}{0}=?$ $a$ or $b$?

Paul
  • 20,553
1

If we take the basic properties of numbers to be true, then certain consequences are forced upon us.

So here's a proof from Spivak's Calculus where I outline what basic properties are being used:

First we prove that for any number a, a⋅0 = 0. 

a⋅0 + a⋅0 = a⋅(0+0), by the distributive law, 
          = a⋅0

So if you add -(a⋅0) to both sides (existence of additive inverses), then it follows that a⋅0 = 0.

Now consider the existence of multiplicative inverses, which states that for every number a ≠ 0, there is a number a-1 such that a⋅a-1 = 1.

Since we have proved that 0⋅a = 0 for all numbers b, then it is clear to see why the restriction of a ≠ 0 must be in place, because we would find ourselves with a number 0-1, which multiplied by 0 gives us 1, a clear contradiction of what we proved (And 0/0 is the same as saying 0⋅0-1).

ginseng_and_honey
  • 11