Why does $\frac {100 - 100}{100 - 100}$ give 3 answers?

Question

I recently saw a Maths Problem, to which I got 3 different answers, all of which seem correct.

Case 1

$\frac {100 - 100}{100 - 100} = \frac{0}{0} = \infty$

Case 2

$\frac{100-100}{100-100} = \frac{(100-100)\div(100-100)}{(100-100)\div(100-100)} = \frac{1}{1}= 1$

Case 3

$\frac{100-100}{100-100} = \frac{10^2 -10^2}{10(10-10)}=\frac{(10+10)(10-10)}{10(10-10)}=\frac{\frac{(10+10)(10-10)}{(10-10)}}{\frac{10(10-10)}{(10-10)}} = \frac{10+10}{10} = \frac{20}{10} = 2$

Now, I know that in mathematics,there is exactly 1 solution to every problem. But then, how can this be?

Answer 1

Both in cases $2$ and $3$, you are dividing by $$100-100=0$$ which is an invalid operation.

In case you are just freaking out, "Where?", I will show it to you.

Case $2$ - Step $2$ and in Case $3$ - Step $4$

Answer 2

It's conventional not to permit division by $0$ but if you set $0=\infty$ then this makes $0$ an absorbing element under both addition and multiplication and that also makes it absorbing under division. Then every version of your equation has a valid solution and the solution by any method gives $\frac {100 - 100}{100 - 100}=0=\infty$.

So you have:

$0a=0=\infty\forall a$

$a\div0=\infty=0\forall a$

and

$a\times0=0=\infty\forall a$

But this has the downside of denying you:

$a+0=a\forall a$ because instead you have $0+a=0=\infty$

so you can do what you're trying to do (divide by zero) but basically it's the equivalent of removing what you conventionally think of as $0$ (the identity under addition) from your algebra and replacing it with $\infty$ (the absorbing element under addition) so it really messes up much of what you take for granted.

Answer 3

I would say, that you cannot divide by zero and therefore the result is undefined.

Actually you divide in all 3 cases by zero.

Step in Case 1:

$\frac{0}{0} = \infty$, where it should be $\frac{0}{0} = undefined$

Step in Case 2:

$\frac{(100-100)÷(100-100)}{(100-100)÷(100-100)} = 1$, where you divide by a term which itself is $undefined$ based on the reason explained above.

Step in Case 3:

$\frac{(10+10)(10-10)}{10(10-10)} = \frac{\frac{(10+10)(10-10)}{(10-10)}}{\frac{10(10-10)}{(10-10)}}$, where you divide numerator and denominator by (10-10) which is zero.

In a practical application, I would prefer to throw an exception with the specifically most convenient reaction.