Why isn't zero divided by zero equal to one?

[Daecon]

Because in nature, nothing could divide itself by nothing all the time.

 

If it made 0/0=1, then bits of stuff would pop into existence throughout the Universe all the time until it was full and there'd be no room left.

 

Unless it already happened, only once, about 13.7 billion years ago...

[DJBruce]

Division is defined as the inverse of multiplication.

 

So if

[math]a \cdot b=c[/math]

Then

[math]\frac{c}{b}=a[/math]

Assume

[math]b=0[/math]

[math]c=0[/math]

[math]a=?[/math]

So if you say

[math]0/0=1[/math]

then

[math]a=1[/math]

But

[math]0 \cdot a=0[/math]

[math]a\neq1[/math]

[math]0 \cdot 3=0[/math]

Because 0 times any number is 0 [math]a[/math] can be any number not just one.

 

 

Also in real life if I have 4 apples and divide them into to 2 equal piles I have 2 piles with 2 apples in each. If I have zero apples and divide them into two equal piles I have to piles with 0 apples in each. If [math]0/0=1[/math] then I would have two piles with 1 apple in one pile and 0 apples in the other.

[alan2here]

I was under the impression that 0/0 = 0 and inf and 1

[DJBruce]

As before lets define division as the inverse of multiplacation so that:

[math]a\cdot b=c[/math]

then

[math]a=c/b[/math]

Assume

[math]a=?

b=0

c=0[/math]

So if

[math]0/0=0[/math]

then

[math]a=0[/math]

But

[math]0\cdot a=0[/math]

and

[math]0\cdot 3=0[/math]

so

[math]a=0[/math] and [math]3[/math]

 

Since this makes no sense and if [math]0/0=0[/math] [math]a[/math] would have to be 0 then we need to assume that this is undefined.

[John Cuthber]

Perhaps the easy answer is that 0/0 can't be 1 because it is 42

0/0=42

Multiply both sides by 0

0=42*0

Simplify

0=0

Perfectly correct.

Unfortunately it works for 0/0= e, or 0/0= -3i or any other number.

Since the ratio of zero to itself can give any answer you like it's fair to say that 0/0 is indeed 1, however it's also 2, 0 or whatever. That's why it's undefined.

[Deja Vu]

What is the answer?

Well, a set of axioms must be established. I think everyone reading can agree that

Axiom 1:

0 = 1/∞

 

Therefore 0/0 identical to (1/∞)/(1/∞). Using (x/y)/(p/m)=(x/y)x(m/p), it must be true that 0/0 is the same as ∞/∞. This is very similar to the first question. Why isn't ∞/∞ simply one?

 

 

No, that's all wrong. 1/∞ is a meaningless expression. It's true that the limit of 1/x as x approaches ∞ is equal to zero, but that doesn't really imply that expression 1/∞ = 0.

 

In any case, the reason 0/0 is undefined is because of all the reasons listed above, you are basically trying to figure out how many times you can put nothing into nothing.

Edited July 20, 2008 by Deja Vu

[Air]

Can somebody please explain to me why zero violates common sense? Every other number including infinity, divided by itself equals one, so why doesn't the same apply to zero as well? After all, since there are numbers which are less than zero, zero itself is logically a number too, and zero does go into itself once. It just seems totally counter-intuitive that 0/0 doesn't equal one.

 

And while we're talking about counter-intuitive math, can someone also explain why a negative times a negative equals a positive, even though a positive times a positive doesn't equal a negative.

 

Please don't use fancy terms that none but a math major could understand.

 

Yes, normally you are told that you get the result 1 if you divide a number by itself.

 

If you permit this result at [math]\frac{0}{0}=1[/math] then the value of all numbers isn't unique anymore.

 

An example:

 

[math]5 \cdot 0 = 9 \cdot 0[/math]. Now divide both sides of this equation by zero:

 

[math]\frac{5 \cdot 0}{0} = \frac{9 \cdot 0}{0}[/math]. According to the assumption you can cancel out the zeros and you'll get

 

[math]5 \cdot 1= 9\cdot 1[/math].

 

Does 5 equal 9? Do you now see why zero is a special case?

 

What is the answer?

Well, a set of axioms must be established. I think everyone reading can agree that

Axiom 1:

0 = 1/∞

 

Therefore 0/0 identical to (1/∞)/(1/∞). Using (x/y)/(p/m)=(x/y)x(m/p), it must be true that 0/0 is the same as ∞/∞. This is very similar to the first question. Why isn't ∞/∞ simply one?

 

Well, ∞ = ∞+∞. Following on, if ∞/∞ was 1, then what is (∞+∞)/∞? Splitting the fraction, it becomes apparent that (∞+∞)/∞ = (∞/∞)+(∞/∞).

 

 

1 can never be equal to 2.

 

[math]\lim_{x \to 0^+} \ \frac 1x=+ \infty[/math]

 

[math]\lim_{x \to 0^-} \ \frac 1x=- \infty[/math]

 

[math]\lim_{x \to +\infty} \ \frac 1x=\lim_{x \to -\infty} \ \frac 1x=0[/math]

 

[math]\frac{\text{Number}}{0}[/math] is undefined [math]\left(\frac{0}{0} \ne \frac{1}{\infty}\right)[/math]. It is different from the limit.

[Severo]

ok try thinking like this

9/3 ; you are asking, how many times does 3 go into 9

8/2 ; how many times does 2 go into 8

 

0/0 ; how many times does 0 go into 0

or how many times does nothing go into nothing. It is not 1 time.

[Josy]

[math]\frac{0}{0}[/math] is an indeterminate form. If we have two functions f and g such that [math]\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x)=0[/math], we have no idea whether [math]\lim_{x\to 0}\frac{f(x)}{g(x)}[/math] tends to a limit, to infinity or not at all. For example, [math]\frac{\sin x}{x}[/math] tends to 1, [math]\frac{x}{x^2}[/math] tends to positive infinity and [math]\frac{\sin{\frac{1}{x}}}{\tan{\frac{1}{x}}}[/math] doesn't do either.

[Zebbygoss]

x/0 = no solution 0/x = 0... 0 is NOTHING. Any number(1) / 0 = 0 but you cant divide zero into anything because it is nothing.. negatives are below zero. They are the same as a something but are the opposite.. If it makes any sense... look at a number line.. -3 -2 -1 0 1 2 3 4.. its not that hard to see this.

Edited September 20, 2008 by Zebbygoss
Correction