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If 0/0= a then 0 X a= 0 is also "well defined".

a/0, for non-zero a, is "undefined"  because if we set a/0= b then a= bX0= 0 which is not true.

0/0 is not defined because if we set 0/0= b then 0= bX0 for any b.

 

Many people say 0/0 is "undetermined" rather than "undefined".  But there is no good reason to set it equal to 0.

25 minutes ago, Country Boy said:

If 0/0= a then 0 X a= 0 is also "well defined".

a/0, for non-zero a, is "undefined"  because if we set a/0= b then a= bX0= 0 which is not true.

0/0 is not defined because if we set 0/0= b then 0= bX0 for any b.

 

Many people say 0/0 is "undetermined" rather than "undefined".  But there is no good reason to set it equal to 0.

Well put. +1

  • 2 weeks later...
  • Author
On 10/27/2021 at 10:33 PM, Country Boy said:

If 0/0= a then 0 X a= 0 is also "well defined".

a/0, for non-zero a, is "undefined"  because if we set a/0= b then a= bX0= 0 which is not true.

0/0 is not defined because if we set 0/0= b then 0= bX0 for any b.

 

Many people say 0/0 is "undetermined" rather than "undefined".  But there is no good reason to set it equal to 0.

I still do NOT think so.

1) 0/0 =0, it is like 1/1=1, there is no such thing, "if we set 0/0= b then 0= bX0 for any b.".

0/0 only has one value, which is 0.

1/1 has only one value, which is 1.

2) It is easy to prove that X/0, X!=0 is NOT defined.

since if X/0 = Y,  then 0 X Y = X, and X!=0 are NOT possible.

 

Edited by PeterBushMan

4 hours ago, PeterBushMan said:

I still do NOT think so.

1) 0/0 =0, it is like 1/1=1, there is no such thing, "if we set 0/0= b then 0= bX0 for any b.".

0/0 only has one value, which is 0.

1/1 has only one value, which is 1.

2) It is easy to prove that X/0, X!=0 is NOT defined.

since if X/0 = Y,  then 0 X Y = X, and X!=0 are NOT possible.

 

Well I don't know what level of mathematics you are operating at but worries about 0/0 are quite common and Country Boy's answer is the simplest available.

4 hours ago, PeterBushMan said:

there is no such thing, "if we set 0/0= b then 0= bX0 for any b."

 

It is difficult to know where to begin to answer when you make this obviously false statement.

What is 4 times zero ?

What is 6 times zero ?

What is 4 times 3 ?

What is 6 times 2 ?

I think that there are a perfectly well defined results for all these multiplications, any many more besides.

In fact, as Country Boy indicated for any or each and every b.

What else do the first two multiplications, taken together show and what do the second two multiplications taken together show ?

This is a unbelievably important result in arithmetic.

Spoiler

They demonstrate (they do not proove ; proof is more difficult.) that the product c of two numbers a, b is also the product of different pairs of numbers, d times e.

 

So to discuss you question of 0/0 without referring to the more fundamental arithmetical operation of multiplication it is necessary to discuss the meaning of

[math]\frac{{A}}{{B}}[/math]

When A and B are mathematical objects.

As an introduction let us consider the value of


[math]\frac{{{x^n} - {a^n}}}{{x - a}}[/math]


Where x is any rational number, and a is any rational constant and n is a power.

Try to calculate the values of the expressions when

x = 5
a = 5
n = 2

and

x = 5
a = 5
n = 3

I make the first one 10 and the second one 75.

Try also to calculate the value of


[math]\frac{{{e^h} - h}}{h}[/math]


When h = 0.

The answer to that one is 1.

 

 

 

 

 

Edited by studiot

Apologies the last expression should have been,

My poor typing again.


[math]\frac{{{e^h} - 1}}{h}[/math]


when h = 0

On 10/27/2021 at 2:03 PM, Country Boy said:

If 0/0= a then 0 X a= 0 is also "well defined".

a/0, for non-zero a, is "undefined"  because if we set a/0= b then a= bX0= 0 which is not true.

0/0 is not defined because if we set 0/0= b then 0= bX0 for any b.

 

Many people say 0/0 is "undetermined" rather than "undefined".  But there is no good reason to set it equal to 0.

Simple and enlightening.

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