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From @fresh_42's Insight

https://www.physicsforums.com/insights/10-math-things-we-all-learnt-wrong-at-school/

Please discuss!

This always sounded to me as if there was obscure mathematics police that forbids us to do so. Why? Well, there is actually a simple reason: ##0## hasn't the least to do with multiplication and even less with division. The question to divide by ##0## doesn't even arise! The neutral element for multiplication is ##1,## not ##0##. That belongs to addition. And there is only one way to combine the two, namely by the distributive law ##a\cdot(b+c)=a\cdot b+b\cdot c.## Let's pretend there was a meaningful way to define ##m=a:0.## Then $$a=m\cdot 0 = m\cdot (1-1)= m\cdot 1 - m\cdot 1= m-m =0$$ So can we at least divide ##a=0## by ##0## and get ##m##? I'm afraid not. Have a look at $$m\cdot a= \dfrac{0}{0}\cdot a = \dfrac{0\cdot a}{0\cdot 1}=\dfrac{0}{0}=m\Longrightarrow a=1$$ or $$2m=(1+1)\cdot m=m+m=\dfrac{0}{0}+\dfrac{0}{0}=\dfrac{0+0}{0}=\dfrac{0}{0}=m$$ and we could only multiply ##m## by ##1##. So even if we define ##m=0/0## we wouldn't get anything useful in the sense that it would be connected to any number we normally use for calculations.

https://www.physicsforums.com/insights/10-math-things-we-all-learnt-wrong-at-school/

Please discuss!

This always sounded to me as if there was obscure mathematics police that forbids us to do so. Why? Well, there is actually a simple reason: ##0## hasn't the least to do with multiplication and even less with division. The question to divide by ##0## doesn't even arise! The neutral element for multiplication is ##1,## not ##0##. That belongs to addition. And there is only one way to combine the two, namely by the distributive law ##a\cdot(b+c)=a\cdot b+b\cdot c.## Let's pretend there was a meaningful way to define ##m=a:0.## Then $$a=m\cdot 0 = m\cdot (1-1)= m\cdot 1 - m\cdot 1= m-m =0$$ So can we at least divide ##a=0## by ##0## and get ##m##? I'm afraid not. Have a look at $$m\cdot a= \dfrac{0}{0}\cdot a = \dfrac{0\cdot a}{0\cdot 1}=\dfrac{0}{0}=m\Longrightarrow a=1$$ or $$2m=(1+1)\cdot m=m+m=\dfrac{0}{0}+\dfrac{0}{0}=\dfrac{0+0}{0}=\dfrac{0}{0}=m$$ and we could only multiply ##m## by ##1##. So even if we define ##m=0/0## we wouldn't get anything useful in the sense that it would be connected to any number we normally use for calculations.

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