# Does 0/0 = a new branch of numbers? Have I made a mistake in the equation? [duplicate]

So I thought...

If $\frac00 = x$... then

$0 = x\cdot0$... then

$0x = 0$... then its technically possible to divide by $0$ again

$\frac{0x}0 = \frac00$ ... since $\frac00 = x$ and $\frac{0x}0 = \frac00$.. then $x = \frac{0x}0$ ...

$0x = 0 ... x = \frac00 ...$

$\frac00 = \frac00=...$ $\frac00$ can only equal itself, meaning the value of it is an exact value, but it cannot equal anything else because the value it has is not from any branch-type of numbers that we know. It isnt imaginary nor anything else. Does this mean we need a new type of number that will make the mechanics of $0$ in maths work?

T.S SVK 1998 (17)

## marked as duplicate by Antonio Vargas, Jeremy Rickard, kjetil b halvorsen, Milo Brandt, muaddibAug 16 '15 at 15:20

• Division by $0$ is not defined. – Wojciech Karwacki Aug 16 '15 at 12:37
• what's the "T.S SVK 1998 (17)" about? it sounds like a weapon – George Aug 16 '15 at 12:41
• I don't think you can really "cross multiply" in this example if that's what you did. Multiplying by zero creates issues. For example if we have the equation $2x=4$ and someone multiples both sides by zero, they often come up $2x(0)=0$ or $0=0$ and come up "all solutions". As we can see this here is an issue which you in some way did in your example. – Ahmed S. Attaalla Aug 18 '15 at 4:46

Well division by two can be defined as $\times2^{-1}$ in other words multiply by the multiplicative inverse of $2$ . To be specific the element $a$ such that $2 \times a=1$ Now such an $a$ is called $2^{-1}$
Can we find an $a$ such that $0 \times a =1$ ?
No because multiplication by $0$ is always $0$. So division by zero is undefined.