Is a Vertical Line the Correct Depiction for this Equation?

[NATE MATE]

Recently, after work beers with a colleague went down a bit of rabbit hole as he attempted to "red pill" me on a couple of topics. One of those happen to relate to the below self-published paper, which was written by someone he went to college with:

https://www.slideshare.net/MrIndererminate/indeterminate-is-not-synonymous-with-undefined

It seems rather compelling on first read, but acknowledging that it is above my paygrade, I decided to search the internet to see if counter arguments had been made, which led me to a thread on this forum, where the paper had been mentioned:

https://www.physicsforums.com/threads/graphing-a-strange-equation.912296/

To ensure I can convey the pro-hole argument effectively, I'm just going to lay out my simple version of it below:

Equality A [ y = (x^2-1)/(x-1) ] cannot be rearranged into Equality B [ y (x-1)=(x^2-1) ] ,as this would involve division by zero.

It is true that graphing B on a cartesian plane will result in a vertical line. However, this does not mean that a vertical line should be depicted when graphing A, as it cannot be rearranged into B without fundamentally changing the equality.

For A, when x=1 y=0/0. Thus hole at co-ordinate (1,2) is the appropriate depiction for A, as all division by zero is undefined.

Is this correct? Are there any other points I should add?

 

[fresh_42]

Equality A [ y = (x^2-1)/(x-1) ] cannot be rearranged into Equality B [ y (x-1)=(x^2-1) ] ,as this would involve division by zero.

It can, if the range of allowed values for $x$ is considered. In detail we have:
$$
A \Longrightarrow B \stackrel{x\neq 1}{\Longrightarrow} A
$$
$A$ has implicitly given that $x\neq 1$ (otherwise the quotient makes no sense), whereas $B$ has not. So in order to get back from $B$ to $A$, $x\neq 1$ has to be added.

It is true that graphing B on a cartesian plane will result in a vertical line.

No.

However, this does not mean that a vertical line should be depicted when graphing A, as it cannot be rearranged into B without fundamentally changing the equality.

The only difference is, that $A$ has a gap at $x=1$ where it is not defined.

For A, when x=1 y=0/0. Thus hole at co-ordinate (1,2) is the appropriate depiction for A, as all division by zero is undefined.

Is this correct? Are there any other points I should add?

Yes, this is correct, only that the line isn't vertical, more diagonal.

 

[Mr Indeterminate]

Equality B absolutely has a vertical line!

It can be graphed on a cartesian plane, by inserting values in for y and then solving for x:

For instance, if y=3 then:

3(x-1)=x²-1

3x-3=x²-1

3x=x²-1+3

3x=x²+2

0=x²-3x+2

0=(x-1)(x-2)

Thus x = 1 or 2 and so the co-ordinates (1,3) and (2,3) are plotted

Repeat this with different values for y and a vertical line in the graph where x=1 will be apparent.

 

[fresh_42]

Equality B absolutely has a vertical line!

It can be graphed on a cartesian plane, by inserting values in for y and then solving for x:

For instance, if y=3 then:

3(x-1)=x²-1

3x-3=x²-1

3x=x²-1+3

3x=x²+2

0=x²-3x+2

0=(x-1)(x-2)

Thus x = 1 or 2 and so the co-ordinates (1,3) and (2,3) are plotted

Repeat this with different values for y and a vertical line in the graph where x=1 will be apparent.

This is completely wrong.

$B$ reads $0=0$ for $x=1$, which doesn't give us anything. For $x\neq 1$ equation $B$ is equivalent to $y=x+1$. So unless you do not introduce another coordinate system, the usual graph is $\{\,(x,y)\,|\,y=x+1\,\}$ which is the line in the image above. It is diagonal and shifted by one up. This is not vertical.

 

[Mr Indeterminate]

This is completely wrong.

$B$ reads $0=0$ for $x=1$, which doesn't give us anything. For $x\neq 1$ equation $B$ is equivalent to $y=x+1$. So unless you do not introduce another coordinate system, the usual graph is $\{\,(x,y)\,|\,y=x+1\,\}$ which is the line in the image above. It is diagonal and shifted by one up. This is not vertical.

Is the method of inserting values for y to determine x not a valid method of graphing equations?

I would read https://www.physicsforums.com/threads/graphing-a-strange-equation.912296/ before you answer that. Numerous senior forum members have agreed in the past that B does indeed have a vertical line (It has never been said that the graph is solely composed of vertical line).

 

[fresh_42]

O.k., the solution $x=1$ is valid, too, as $B$ is a tautology in this case. So you are right, the graph of solutions is

Sorry.

 

[Mr Indeterminate]

Alright now that we are on the same page on that one. I'm going to come out and say that I don't agree with NATE MATE's argument.

For equation B, if x=1 then y= 1 or 2 or 3 or -4 or 0.34928 or -200 ... basically its any number. Merriam-Webster defines indeterminate as "Having an infinite number of solutions". Thus, I'm inclined to conclude that for equation B if x=1 then y=Indeterminate.

 

[fresh_42]

If $x=1$ then $\forall\, y\, : \,y(x-1)=x^2-1$. I don't think we can call $y$ indeterminate. $y$ cannot be determined, yes, but the reason is, that $y$ is simply not part of the statement anymore, since $y\cdot (1-1) =0$ and $B$ is reduced to $x=1\,\wedge \,0=0$ as part of the assumption to $B$ in this special case. It's somehow as if we said, that $z$ is indeterminate. So without an occurrence, can you attribute an adjective? The statement $B$ is probably better called a tautology and $y$ cannot be referenced anymore.

If $x\neq 1$ then $y=x+1$ and $y$ is determined.

So $B$ determines $y$ in some cases, and does not determine $y$ in another case. But that doesn't make $y$ indeterminant, since it vanished!

 

[Ray Vickson]

Is the method of inserting values for y to determine x not a valid method of graphing equations?

I would read https://www.physicsforums.com/threads/graphing-a-strange-equation.912296/ before you answer that. Numerous senior forum members have agreed in the past that B does indeed have a vertical line (It has never been said that the graph is solely composed of vertical line).

The formula $f(x) = (x^2-1)/(x-1)$ defines a valid mathematical object only if $x \neq 1$. It describes a perfectly well-defined mathematical function on the open domain $(-\infty, 1) \cup (1, \infty),$ which is the real line with the value $x=1$ missing. Of course, on that domain we have the simpler version $f(x) = x+1$, and that form can be extended to the whole line, including the point $x=1$.

If you want to plot a graph of $y = f(x)$ by choosing $y$ and then solving for $x$ you will find that the value $y=2$ is missing: there is no solution of the equation $2 = (x^2-1)/(x-1)$. Of course, there is a solution of $2 = x+1$, but that is a different problem for a different function.

I think one needs to be careful to distinguish between "no solution" and "non-unique solution = vertical line". To a certain extent, we can argue that it is a matter of taste: I prefer to go along with standard mathematics that regards---for excellent reasons--- division by zero to be always disallowed. Others, who may have a more informal and flexible set of mathematical rules, will regard something like 0/0 to be whatever they want.

 

[Mr Indeterminate]

The formula $f(x) = (x^2-1)/(x-1)$ defines a valid mathematical object only if $x \neq 1$. It describes a perfectly well-defined mathematical function on the open domain $(-\infty, 1) \cup (1, \infty),$ which is the real line with the value $x=1$ missing. Of course, on that domain we have the simpler version $f(x) = x+1$, and that form can be extended to the whole line, including the point $x=1$.

If you want to plot a graph of $y = f(x)$ by choosing $y$ and then solving for $x$ you will find that the value $y=2$ is missing: there is no solution of the equation $2 = (x^2-1)/(x-1)$. Of course, there is a solution of $2 = x+1$, but that is a different problem for a different function.

I think one needs to be careful to distinguish between "no solution" and "non-unique solution = vertical line". To a certain extent, we can argue that it is a matter of taste: I prefer to go along with standard mathematics that regards---for excellent reasons--- division by zero to be always disallowed. Others, who may have a more informal and flexible set of mathematical rules, will regard something like 0/0 to be whatever they want.

So with equation A $y=(x^2-1)/(x-1)$ you say a hole is the appropriate depiction.

Do you agree that equation B $y(x-1)=(x^2-1)$ has a vertical line?

 

[Ray Vickson]

So with equation A $y=(x^2-1)/(x-1)$ you say a hole is the appropriate depiction.

Do you agree that equation B $y(x-1)=(x^2-1)$ has a vertical line?

That does not really matter, since that is for a different problem. When trying to go from A to B you are essentially saying that when you write
$$(x-1)y = \frac{x^2-1}{x-1} (x-1) \hspace{4ex}(1)$$
you want to cancel the $(x-1)$s in the numerator and denominator on the right and so be left with just $x^2-1$. That cancellation is valid for all $x \neq 1$, but when $x=1$ it is pulling the old $0/0 = 1$ trick, which is invalid for main-stream mathematics.

 

[Mr Indeterminate]

That does not really matter, since that is for a different problem. When trying to go from A to B you are essentially saying that when you write
$$(x-1)y = \frac{x^2-1}{x-1} (x-1) \hspace{4ex}(1)$$
you want to cancel the $(x-1)$s in the numerator and denominator on the right and so be left with just $x^2-1$. That cancellation is valid for all $x \neq 1$, but when $x=1$ it is pulling the old $0/0 = 1$ trick, which is invalid for main-stream mathematics.

Ok, so what your saying is that equation A cannot be rearranged into equation B.

However, if you were just to start with equation B, would it be depicted with a vertical line?

 

[Mark44]

Ok, so what your saying is that equation A cannot be rearranged into equation B.

I don't think that's what Ray Vickson was saying. Starting with equation A, $y = \frac{x^2 - 1}{x - 1}$, you can get to equation B by multiplying both sides by x - 1. However, equation A is undefined if x = 1, so 1 is not in the domain of the function represented by this equation. As already noted, the graph of equation A is a a line with slope 1 and y intercept at (0, 1), but with a removable discontinuity at (1, 2).

However, if you were just to start with equation B, would it be depicted with a vertical line?

The graph of equation B would be two intersecting straight lines.
If x = 1, the equation simplifies to $y \cdot 0 = 0$, which is true frue for all real y. If $x \ne 1$: the equation simplifies to y = x + 1, as is shown in post #6.

Clearly the graphs are different, which means that the two equations aren't equivalent.

 

[Mark44]

There's really not any more to be said, so I'm closing this thread.