# The difference between 0/0 from 1/0

ziemel_dlam
“The result of any division by zero is undefined.”

Although I agree with the above statement, the purpose of this thread is to highlight a fundamental distinction between 0/0 and 1/0.

To highlight this difference between 0/0 and 1/0 it requires approaching division with what I call the reversal perspective. This perspective considers division to be the reversal of multiplication, specifically:

A x B = C then C/B = A.

If we consider that 0/0 = A with A being the unknown then A x 0 = 0. As soon as you start to insert numbers in for A, for instance 1 x 0 = 0 or 534 x 0 = 0, all of them hold. 0/0 is undefined because the answer could potentially be any value.

Now when we look at 1/0, 534/0 or any other value divided by 0, there is a solid distinction:

A x 0 = 1 and A x 0 = 534 are invalid, A x 0 always = 0

The reality is that 1/0 is undefined because there are no values to which it equals; in complete contrast to 0/0 which could be any value.

Although this distinction is clear in the above rationale, it has been overlooked when graphing functions. When graphing y= (x^2-1)/x-1 mainstream mathematics states that the graph is equivalent in appearance to the function y=x+1, with the exception of when x=1 where it is equivalent to 0/0. As 0/0 is considered to be undefined in the same way that 1/0 is, the mainstream depiction is of a hole on graph at that point.

In consideration of the highlighted distinction between 0/0 and 1/0, you would think that the hole would be more suited to 1/0 and that 0/0 would be better represented by a vertical line. Simply because 0/0 could result in any of the values on the number line, whereas 1/0 is not equal to any value on the number line.

As some may not be convinced by the rationale presented or convinced that a vertical would be more appropriate in a graph where a point equals 0/0. I have come up with the horizontal line test.

The rationale of this test is simple. When you have two equations equal each other and solve for an unknown variable, it allows you to determine the intercept point on a graph. One way to plot an equation is to have them equal to a mass of different horizontal lines, such as y = 1 or y = 2 and then mark the intercepts on a Cartesian plane. So lets to the test with y = 3, y = 2 and y = 1 and see where the intercepts are.

Where does y = 3 and y =(x^2-1)/x-1 intercept?
(x^2-1)/x-1 = 3
x^2-1 = 3x-3
0 = x^2-3x+2
0 = (x-2)(x-1)
their for x= 2 and 1
Intercepts detected at the points (1,3) and (2,3)

Where does y = 2 and y =(x^2-1)/x-1 intercept?
(x^2-1)/x-1 = 3
x^2-1 = 2x-2
0 = x^2-2x+1
0 = (x-1)(x-1)
their for x= 1
Intercept detected at the point (1,2)

Where does y = 1 and y =(x^2-1)/x-1 intercept?
(x^2-1)/x-1 = 1
x^2-1 = x-1
0 = x^2-x
0 = x(x-1)
their for x= 0 and 1
Intercepts detected at the points (0,1) and (1,1)

Even if you don't mark out each of the intercepts on a Cartesian plane, its rather obvious that there was intercepts detected every time a horizontal line passed though x = 1. this would be consistent with their being a vertical line at x = 1 and completely inconsistent with a hole being present at the point (1,2).

If you agree there is a difference between 0/0 and 1/0, then do pass this information onto your math buddies. If you don’t think there is a difference, please elaborate your reasoning in the comments.

ヽ༼ຈل͜ຈ༽ﾉ Thanks for reading ヽ༼ຈل͜ຈ༽ﾉ

## Answers and Replies

Homework Helper
Gold Member
Dearly Missed
1. We call expressions like 0/0 indeterminate, rather than "undefined". Major difference!
Still, it doesn't qualify as a number.

2. " it has been overlooked when graphing functions. "
Nope. Because a function is DEFINED to be such that for each x, there is a UNIQUE number assigned to it called f(x). Thus, wherever there isn't any functional relationship between the x and the y, the function is, quite simply, NOT defined, and the graph of that function has a hole there.

Staff Emeritus
Although I agree with the above statement, the purpose of this thread is to highlight a fundamental distinction between 0/0 and 1/0.

To highlight this difference between 0/0 and 1/0 it requires approaching division with what I call the reversal perspective. This perspective considers division to be the reversal of multiplication, specifically:

A x B = C then C/B = A.
That is precisely how division is defined if B is not zero. C/B=A is equivalent to saying that the A is unique quantity such that AB=C.

What if B is zero? If C isn't zero there is no quantity A such that A*0=C. Dividing a non-zero quantity by zero is an undefined operation. On the other hand, if B is zero then any and every value for A satisfies A*0=0. Dividing zero by zero is not undefined. It is instead indeterminate. That's what arildno already said.

In other words, that there is a fundamental distinction between 0/0 and 1/0 is well-known.

Homework Helper
Gold Member
Dearly Missed
As for the graph with a vertical line (which is a perfectly legit. mathematical object), it is NOT the graph of a function.

It can be thought of as the UNION set of the graphs of (uncountably many) functions, each being defined, and distinguished by the other through the value specified on the relevant vertical line.

Mentor
Where does y = 3 and y =(x^2-1)/x-1 intercept?
(x^2-1)/x-1 = 3
x^2-1 = 3x-3
0 = x^2-3x+2
0 = (x-2)(x-1)
their for x= 2 and 1
Intercepts detected at the points (1,3) and (2,3)
You wrote (x^2-1)/x-1, which means ##\frac{x^2 - 1}{x} - 1##.
Obviously, what you meant was ##\frac{x^2 - 1}{x - 1}##.
To convey that meaning when you write the expression on a single line, put parentheses around both numerator and denominator, like so - (x2 - 1)/(x - 1).

A more serious problem is that you are ignoring the domain for your rational expression. Since division by zero is not defined, when you multiply both sides of the equation by x - 1, it should be understood that x ≠ 1.

This means that the only solution of the equation (x2 - 1)/(x - 1) = 3 is 2.
Where does y = 2 and y =(x^2-1)/x-1 intercept?
(x^2-1)/x-1 = 3
You mean (x2 - 1)/(x - 1) = 2.
As before, it must be assumed that x ≠ 1.
x^2-1 = 2x-2
0 = x^2-2x+1
0 = (x-1)(x-1)
their for x= 1
Intercept detected at the point (1,2)
Since x cannot be 1, there is no solution to the equation.
Where does y = 1 and y =(x^2-1)/x-1 intercept?
(x^2-1)/x-1 = 1
x^2-1 = x-1
0 = x^2-x
0 = x(x-1)
their for x= 0 and 1
Intercepts detected at the points (0,1) and (1,1)
No. As in the previous two examples, x cannot equal 1. So the only solution is x = 0.
Even if you don't mark out each of the intercepts on a Cartesian plane, its rather obvious that there was intercepts detected every time a horizontal line passed though x = 1. this would be consistent with their being a vertical line at x = 1 and completely inconsistent with a hole being present at the point (1,2).
Nope, don't buy it. As I have shown, in each example where you claim there are two points of intersection, there was actually only one.
If you agree there is a difference between 0/0 and 1/0, then do pass this information onto your math buddies. If you don’t think there is a difference, please elaborate your reasoning in the comments.
Yes, I agree that there is a difference between 0/0 -- one of several indeterminate forms -- and 1/0 -- just plain undefined.

ziemel_dlam
: O thanks Mark 44 for pointing out the errors.

But I must say that your argument against the reasoning is kinda strange, in general terms 'proof against conjecture false due to conjecture' does quite stand up as reasoning.

As arildno has informed me that 0/0= indeterminate. Why isn't a vertical line the better way to depict indeterminate?

My reasoning is:
1. it would be better to differentiate between 0/0 & 1/0 clearly when graphing
2. It makes logical sense that if something could be equal to that value than it will be marked on the graph
3. when you stop assuming its a hole and check what is there, you find a line

aww yea arildno, its a relation sorry

Why isn't a vertical line the better way to depict indeterminate?

Example: graph the continuous extension of sin(x)/x. In this case, we obtain a 0/0 indeterminate form at x=0, but we will graph (0,1) and not a vertical line.

1. it would be better to differentiate between 0/0 & 1/0 clearly when graphing

Why? You need to actually make a case.

2. It makes logical sense that if something could be equal to that value than it will be marked on the graph

'Could' is not a strong enough reason, unless that is exactly what the question is asking for to begin with.

3. when you stop assuming its a hole and check what is there, you find a line

Or a point (i.e. removable singularity) or nothing (i.e. essential singularity).
0/0 means you haven't finished your analysis.

ziemel_dlam
If I put this entire thread down to one question

'If 1/0 = undefined and 0/0 = indeterminate, why represent them on a graph the same?'

Does it help?

Staff Emeritus
If I put this entire thread down to one question

'If 1/0 = undefined and 0/0 = indeterminate, why represent them on a graph the same?'

Does it help?
No. Do these look at all like they are represented the same? http://www.sagemath.org/calctut/pix-calctut/inflimits01.png [Broken]

Last edited by a moderator:
Homework Helper
Gold Member
Dearly Missed
If I put this entire thread down to one question

'If 1/0 = undefined and 0/0 = indeterminate, why represent them on a graph the same?'

Does it help?

As DH shows, we don't.
The reason why we place a HOLE there is that we cannot assign, by any reasoned argument, a UNIQUE function value there. Thus, the function is undefined there.
While 0/0 is called "indeterminate" and 1/0 "undefined", BOTH cases are "equally bad" for the FUNCTION, and the FUNCTION is as undefined at a 0/0 position as it is on a 1/0 position. (A function REQUIRES, by definition, that a unique value can be assigned to the x-input, in the 1/0-case, no such value (number) exists, in the 0/0-case, the UNIQUENESS criterion fails)

The graph with a vertical line through it is, quite simply, not a graph of a function, and that's why we don't draw it.

Last edited:
Mentor
But I must say that your argument against the reasoning is kinda strange, in general terms 'proof against conjecture false due to conjecture' does quite stand up as reasoning.
How is my argument against your reasoning strange? In your three examples using y = (x2 - 1)/(x - 1), you claimed that the line y = 3 intersected the graph of the rational function twice, that the line y = 2 intersection the rational function graph once, and the line y = 1 intersection the graph of the rational function twice.

In actual fact, the lines y = 3 and y = 1 intersect the graph once, and the line y = 2 doesn't intersect the graph at all. What you were doing, and I showed this, was multiplying both sides by zero (when x = 1), but your work didn't reflect that fact.

ziemel_dlam said:
'proof against conjecture false due to conjecture'
- I don't have any idea what you're saying here.

ziemel_dlam said:
does quite stand up as reasoning
Did you mean "does not"?

Mentor
This thread is finished...