Can 0/0 Ever Equal 1?

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SUMMARY

The discussion centers on the mathematical concept of 0/0 and its implications. Participants clarify that division by zero is undefined, and specifically, 0/0 is termed "undetermined." They explore functions like y=x/x and y=x/x^3 to illustrate that while 0/0 could suggest a value of 1, it leads to contradictions and undefined behavior in calculus. Ultimately, the consensus is that 0/0 cannot equal 1 due to the inherent flaws in the reasoning surrounding division by zero.

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  • Understanding of basic arithmetic operations, particularly division.
  • Familiarity with mathematical functions and limits.
  • Knowledge of calculus concepts, especially indeterminate forms.
  • Comprehension of real number axioms and properties.
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  • Study the concept of limits in calculus, particularly indeterminate forms.
  • Learn about the implications of division by zero in mathematical functions.
  • Explore the definitions and properties of real numbers and their axioms.
  • Investigate the role of limits in determining the behavior of functions approaching 0/0.
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foges
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I did a search first but didnt really find anything, this is something I've been wondering for a bit:

Both my calculator and math teacher tell me I can't ever divide by zero, but what if you had 0/0 ? couldn't that work, and equal 1? or is there a case in which this would not work?
 
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Division by 0 is not defined. It's tempting to say that since every other number divided by itself is 1, so should 0. But this would have some strange consequences. Consider the function
y=\frac{x}{x}
If x is not 0, then y=1. So it seems natural that y should also be 1 at x=0. But now consider this function:
y=\frac{x}{x^3}
When x=0, we also get y=0/0. But when x gets close to 0, y gets larger and larger and approaches infinity. It doesn't seem reasonable to say y=1 when x is zero. This is one reason division by 0 is left behind. Functions that approch 0/0 are important in the study of calculus.
 
In fact, given any real number r, we can find fractions f(x)/g(x) where both f and g have limit 0 (as x goes to a) but f(x)/g(x) goes to r.

By the way, while we say that a/0, for a not 0, is "undefined", it is common to say that 0/0 is "undetermined". If we try to set a/0= x then we must have a= 0(x) which is impossible. On the other hand, if 0/0= x then we must have 0= 0(x) for which is true for all values of x.
 
Remember that EVERY fraction a/b can be seen as the product a*(1/b)

The number (1/b) is that number which multiplied with b yields 1, i.e, b*(1/b)=1, by definition of 1/b.
1/b is called the reciprocal of b.

However, without resorting to any idea of reciprocals at all, we may prove that for ANY number a, we have 0*a=0.
But that means that THERE CANNOT EXIST A NUMBER 1/0!

Therefore, the reciprocal 1/b can only be defined for numbers not equal to zero.

Thus, the expression 0/0=0*1/0 tries to do the impossible thing, namely multiplying together something that IS a number (0), and something that ISN't a number (1/0). But multiplication requires that both factors are, indeed, numbers..
 
Thanx, i guess i was thinking too simply. I managed to dissprove myself:

0=0*3

0/0 = (0*3)/0 = (0/0)*3

And if 0/0 = 1, Then:

1 = 1*3

1=3
 
I never understood a thing... isn't right to say that:

\frac{0}{0}=\mathbb{R}

When i can't solve a problem, can i go to the prof saying:" the solution is 0/0!"?
 
R is a set. Are you asserting that, what ever 0/0 is, it is a set?
 
matt grime said:
R is a set. Are you asserting that, what ever 0/0 is, it is a set?
yes. for example, the limits in the indeterminate form, if it can be removed, they can assume any result in R you wish. or not?
 
In this case, you have devised a fantasy function having two real arguments going to some set of sets. Nothing wrong with that of course, except it hasn't anything to do with a BINARY OPERATION, like multiplication or division.
 
  • #10
Sure you could choose to have a function where you choose whatever value from the real numbers you want, but you will find picking the right number is pretty hard :P
 
  • #11
foges said:
Thanx, i guess i was thinking too simply. I managed to dissprove myself:

0=0*3

0/0 = (0*3)/0 = (0/0)*3

And if 0/0 = 1, Then:

1 = 1*3

1=3
There's a problem in your second step =).

It'd be (0*3)/(0*3). Even then, it'd be 0 in the end.

But it's just fantasy math with a lot of problems/flaws in the reasoning heh.
 
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  • #12
AngeloG said:
There's a problem in your second step =).

It'd be (0*3)/(0*3). Even then, it'd be 0 in the end.

But it's just fantasy math with a lot of problems/flaws in the reasoning heh.

It doesn't matter if they are equivalent, since it is obviously the case that 0*3=0.
 
  • #13
Then you have the case of saying 0/0 = 1; when before you stated 0 = 0*3... Where in you have 0/0 = 3 :p. A lot of flaws.

You can't switch things up like that just because you feel like it.
 
  • #14
AngeloG said:
Then you have the case of saying 0/0 = 1; when before you stated 0 = 0*3... Where in you have 0/0 = 3 :p. A lot of flaws.

Obviously there are flaws. Was that unexpected? Division by zero is not defined. 0=0*3 is a common fact that follows from axioms for the real numbers, nowhere did we say 0/0=3, but this "disproof" is based on an assumption that 0/0=1.

AngeloG said:
You can't switch things up like that just because you feel like it.

If they are EQUAL you certainly can. You seem to be saying something along the lines of I can't interchange 22 and 4. despite the fact that they are the same thing.
 
  • #15
The thread has seem to run its course, so I'm closing it.
 
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