I just want to make sure about a concept : What is #0/0#? Usually anything divided by #0# is undefined, but a past student of mine reckons they were taught it is #1#. Is there any theorem or proof of this? Thanks

5 Answers
Mar 28, 2016

Your student either mis-remembers or was mis-taught. (Both things happen.)

Explanation:

If we attempt to make #0/0# equal to one (by definition or whatever), then we lose our number system.

Using what we know about multiplication, we can prove that there is only one number.

If #0/0 = 1#, then, using the usual definition of multiplication, we get

#2xx0/0 = (2xx0)/0 = 0/0 = 1# and also

#2xx0/0 = 2xx1 = 2#, so we can prove that #1=2#. This is not a useful result.

In fact we can use any number #x# in place of #2# to show that: if #0/0=1# (and we keep our definition of multiplication) then #x=1#.

In fact any attempt to define division by #0# for any numerator will result in destroying the number system.

In fact we cannot even say that limits of the form #0/0# evaluate to #1# unless we are prepared to lose tangent lines and rates of change. Limits of difference quotients would all evaluate to #1#.

Mar 28, 2016

#0/0# is undefined

Explanation:

Jim has covered this quite well, so I will add little more.

Here are some spurious proofs for illustration/consideration:

#color(white)()#
"Proof" 1

#x/x = 1# for any number #x#, so surely #0/0 = 1# for consistency.

Division by #0# is always undefined.

#color(white)()#
"Proof" 2

#x^0 = 1# for any number #x#

So #x/x = x^1 * x^(-1) = x^(1-1) = x^0 = 1# for any #x#, in particular #0#

This is a thinly disguised attempt to justify #0/0 = 1# using the convention that #0^0 = 1#, just like #x^0 = 1# for any #x != 0#

Mar 29, 2016

The question asks to prove/disprove something which was never defined.

Explanation:

This is how I was taught.

The problem has been posed with the assumption that dividing by 0 is a legitimate operation.

By definition, division operation is opposite of multiplication operation. e.g.,

If #c# times #b# equals #a#, can be written symbolically as

#c times b = a#
then #a# divided by #b# equals #c#, can be written as

#a/b = c# for all values of #b# except for #b=0#.
We must remember that #a, b and c# are unique numbers.

Division with 0 was never defined.

Moreover, the answer resides in the question itself.
Usually anything divided by 0 is undefined

and 'anything' includes all numbers including 0

The ancient Samskrit text which is treated as definition of zero also did/does not talk of division by zero.
enter image source here

Mar 31, 2016

I agree that #0/0# is undefined
but it raises the question of whether it is (arbitrarily) definable.

Explanation:

If division is defined as the opposite of multiplication
so that #a div b = c# means #cxxb=a#
then
#color(white)("XXX")a div 0# for #a!=0#
is quite different from
#color(white)("XXX")a div 0# for #a=0#

If #a!=0# then there is no possible value that could be defined for #c# such that
#color(white)("XXX")cxx0=a#
however if #a=0# we could define #c# to be some (perhaps arbitrary) value and maintain consistency.

Sep 14, 2016

#0/0# is undefined

Explanation:

#0/0# is undefined..

One could explain this by saying there are 3 rules in place here.

"Zero divided by anything is equal to 0"

#0/5 =0," "0/25 = 0, " "0/(-14) = 0# etc

"Anything divided by itself is equal to 1.

#5/5 = 1, " "37/37 = 1, " " (-12)/(-12) = 1# etc

"Division by 0 is not permissible/undefined"
Dividing by 0 actually gives infinity as the answer, but infinity is not a number.

So, we have 3 possible answers using valid maths concepts.

So which is it? #"Is" 0/0 =0?," Is" 0/0 = 1?" Is" 0/0= "infinity"?#

No-one knows, so it best to just say that #0/0# is undefined.

Please refer to the link given below/