# A problem with 0/0=0 [duplicate]

So what is the problem/contradiction If I state I can divide by 0 but the number will be 0 I tried a few thing but didn't find a problem

• Division is best understood in terms of multiplication since that's exactly where it comes from. The quantity $\frac{a}{b}$ is the unique number $x$ such that $a = bx$. Is there a unique number $x$ such that $0 = 0x$? – Cameron Williams Nov 26 '19 at 15:43
• When we say $a/b = c$ , we mean we have to find a number $c$ such that when multiplied by $b$ , would give us $a$. In case of $0/0$ , we need a number $x$ such that $x\times0 = 0$, but it is true for all values of $x$. – The Demonix _ Hermit Nov 26 '19 at 15:44
• Note that $\frac 0 0 =c \implies 0=c\cdot 0=0$ for any $c$. – user Nov 26 '19 at 15:44
• Not clear... "divide by $0$" has no meaning: this is why $\dfrac 0 0$ is undefined. We divide by two when we cut in a half; we divide by one when we do not divide at all. Thus, there is nothing "less than" not divide at all. – Mauro ALLEGRANZA Nov 26 '19 at 15:45
• $\frac{0}{0}+\frac{1}{1}$ is what then? If addition worked for rationals as it should usually, noting that $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$ we would have had $1 = 0+1=\frac{0}{0}+\frac{1}{1}=\frac{0\cdot 1+0\cdot 1}{0\cdot 1} = \frac{0}{0}=0$ – JMoravitz Nov 26 '19 at 15:45

Let $$\frac 0 0=0$$

then $$\frac 0 0+\frac ab=0+\frac ab=\frac ab$$

However $$\frac 0 0+\frac ab=\frac {a\cdot 0+b\cdot 0} {0\cdot b}=\frac 0 0$$

Which implies $$\frac ab=0$$.

Let $$a=2$$ and let $$b=1$$.

Contradiction , as it implies $$2=0$$.

Furthermore, Consider the group of integers defined w.r.t multiplication. In any group, there exists only a single identity element and in the group of integers /0. It is 1. However by introducing 0*0^(-1)=0, it implies 0 is the identity element. This would also mean that there are 2 multiplicative identities for a single group. The whole group theory will be in big trouble as a result and we would get a lot of contradictions too.