certainly,that is not about 1 by 0.....it is \[\frac{a}{0}=infinity\]
where a is non-zero real number.(you will know later why i am writing non-zero).
here i found something a few weeks ago.let a infinite sequence
$x+x+x+x+x+x+x..........$
x is non-zero real number.suppose the sum of the sequence is $S$.
$S=x+x+x+x+x.......$
consider this sequence as a geometric sequence where 1st term,$a=x$,ratio,$r=1$,we know if $-1<r<1$,a as 1st term,then the sum is \[\frac{a}{1-r}\]
theoratically,(applying the theory, though $r=1$)
\
but,$S=infinity$,(of course,as it is sum of infinite $x$)
so,\[\frac{x}{0}=infinity\]
can we call it a proof?
now consider $x=0$
that means $S=0$ (why not? sum of infinite 0 is 0)
but theoratically like above,
\
so,\[\frac{0}{0}=0\]
that's why i wrote non-zero.
do you find anything wrong?then write it.i guess everything is ok.do you know any proof of that,infinity?please,post ,if you know.....
[sorry,did not use here the infinite sign,actually i don't know how to write it]
And ur prof contradicts with principles of mathematics.Look, if\[a/0=b\]
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