Explaining the Vertical Line Feature in Graphing Equations

[Ned Eterminita]

Hello Forum,

I've gone about graphing the below equation by inserting values in for y and then solving for x:

y(x-1)=x²-1

For instance, I say y=3 and then solve:

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 and 2 and so I plot co-ordinates (1,3) and (2,3).

Now I repeated this quite a few times with different values and it appears that there is a vertical line in the graph where x=1.

Can anyone explain why this vertical line feature appears in the graph?

 

[FactChecker]

At x=1, any value of y will satisfy the original equation. The original equation at x=1 is y×0=0. So any value of y will work.

 

[Mark44]

Hello Forum,

I've gone about graphing the below equation by inserting values in for y and then solving for x:

y(x-1)=x²-1

You've gone to a lot of effort below that is unnecessary. Divide both sides by x - 1 to get
$y = \frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1$, if $x \ne 1$.

For instance, I say y=3 and then solve:

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 and 2 and so I plot co-ordinates (1,3) and (2,3).

Your first point is incorrect. If x = 1, the equation you started with becomes y * 0 = 0, which, as already stated, can't be solved for y.

Now I repeated this quite a few times with different values and it appears that there is a vertical line in the graph where x=1.

Can anyone explain why this vertical line feature appears in the graph?

There is no vertical line. What you have in your equation is a removable discontinuity, a single point at which the equation is not defined -- at (1, 2). For any other value of x, a y value can be found.

 

[mfb]

Your first point is incorrect. If x = 1, the equation you started with becomes y * 0 = 0, which, as already stated, can't be solved for y.

OP doesn't solve for y. Finding possible x-values for a given value of y is perfectly possible. It is not the most efficient way, but it is not wrong.

There is no vertical line.

Of course there is. x=1 is a solution for all y. The equation is perfectly defined at x=1. It just reads 0=0, which is true. No one said you have to be able to solve for y.

 

[Mark44]

OP doesn't solve for y. Finding possible x-values for a given value of y is perfectly possible. It is not the most efficient way, but it is not wrong.

Finding possible x-values is perfectly possible, but I can't help wondering if we're seeing later work in the problem, and not the original problem, such as if it were "sketch a graph of $y = f(x) = \frac{x^2 - 1}{x - 1}$."

Of course there is. x=1 is a solution for all y. The equation is perfectly defined at x=1. It just reads 0=0, which is true. No one said you have to be able to solve for y.

Right, but again, this might be the answer to the question as posed in this thread, but not the one really intended.

To solve for x in the equation $y(x - 1) = x^2 - 1$ you would need to get all the terms involving x on the same side, by dividing both sides by x - 1. One would have to stipulate that x could not be 1, for obvious reasons. That would eliminate the possibility of an infinite number of y values when x = 1.

 

[mfb]

To solve for x in the equation $y(x - 1) = x^2 - 1$ you would need to get all the terms involving x on the same side, by dividing both sides by x - 1. One would have to stipulate that x could not be 1, for obvious reasons. That would eliminate the possibility of an infinite number of y values when x = 1.

To solve for x, the correct answer is "x=1 or x=y-1".

While we can speculate if OP's question was the original problem statement, if you answer a different question than the question that got asked, I think you should at least point out that you are not answering the question that was asked.

 

[Mark44]

if you answer a different question than the question that got asked, I think you should at least point out that you are not answering the question that was asked.

I agree. However, we get many questions that stem from incorrect work or are poorly posed, so a bit of interpretation is called for.

@Ned Eterminita, was the question you posted the original problem statement?

 

[Ned Eterminita]

@Ned Eterminita, was the question you posted the original problem statement?

This is how the question was posed:

Yo Cuz,

You want to make $20?

Graph this equation [ y(x-1)=x^2-1] and then explain its unusual feature. If you do it correctly I'll give you $20.

Look forward to seeing you and your graph at Tony's wedding next Saturday.

Ciao

So yea, I've got a very strange cousin. He's always been into riddles, paradoxes and generally thinking about things too deeply. While I'm not someone to do math in my free time, I will do it for money.

So at first, I went about moving (x-1) to the right hand side and then simplifying to y=x+1. This produces no unusual features. I thought about it for a while and realized that I made the mistake of assuming that 0/0 = 1.

On second approach I've gone about inserting values for y and then solving for x, this way I avoid division by zero. This time I got the vertical line, which makes sense considering that y can be anything for y*0= 0.

So what do you recon guys, will I be getting that $20?

P.S. Thanks for the help

 

[Aufbauwerk 2045]

Hint:

Think about f(x) = x^2 and g(x) = x-1. What happens when you compose these functions? Try f(g(x)) and g(f(x)). The order does matter.

I mention composition because learning how to compose and decompose functions is a very important skill.

 

[Ned Eterminita]

Hint:

Think about f(x) = x^2 and g(x) = x-1. What happens when you compose these functions? Try f(g(x)) and g(f(x)). The order does matter.

I mention composition because learning how to compose and decompose functions is a very important skill.

Hi Aufbauwerk, I can't make sense of your approach to the question.

 

[Aufbauwerk 2045]

Sorry, it was not a good hint because I was tired.

I will make up for it by giving the correct explanation for you.

We have f(x) = y = (x^2 - 1)/(x-1).

Looks like it has a discontinuity at x = 1, because the denominator = 0.

But the problem goes away when you factor the numerator, so y = f(x) = (x+1)(x-1)/(x-1) = x + 1.

In fact it's just a straight line. f(1) = 2.

[EDIT Sorry, I was a bit careless. I should say lim f(x) as x -> 1 = 2. I've only seen this problem in the context of doing the limit.]

 

[FactChecker]

Sorry, it was not a good hint because I was tired.

I will make up for it by giving the correct explanation for you.

We have f(x) = y = (x^2 - 1)/(x-1).

Looks like it has a discontinuity at x = 1, because the denominator = 0.

But the problem goes away when you factor the numerator, so y = f(x) = (x+1)(x-1)/(x-1) = x + 1.

In fact it's just a straight line. f(1) = 2.

The fraction is undefined at x=1. The original statement of the OP is fine as long as we realize that any value of y satisfies y⋅0 = 0. That is why the OP finds a vertical line at x=1.

 

[Aufbauwerk 2045]

The fraction is undefined at x=1. The original statement of the OP is fine as long as we realize that any value of y satisfies y⋅0 = 0. That is why the OP finds a vertical line at x=1.

Yes I got sloppy. Sorry. I'm thinking of the limit of f(x) as x goes to 1. I thought that was the question.

But if you want to graph the original function just put a little circle at f(1) to indicate a discontinuity. I'm not sure what's the point, if we are not talking about the limit.

 

[Aufbauwerk 2045]

P.S. drawing the straight line x=1 to indicate the discontinuity (due to the function f(x) being undefined for x=1) is not correct. We just draw a little circle to indicate that at one point only the function is undefined. At least that's the way it's done in the textbooks.

 

[Aufbauwerk 2045]

This actually raises an interesting, or maybe a silly, question. It seems very basic.

Suppose we start with the function f(x) = x + 1. This is defined for all x and is a straight line.

Now we multiply both numerator and denominator by x - 1. So we get the function of the problem, f(x) = (x+1)(x-1)/(x-1) which is undefined for x = 1.

This seems nonsensical. Is it even valid to multiply by (x-1)/(x-1)? If it is, why isn't it equally valid to cancel the x-1 terms?

In other words, it seems strange that we can use algebra to transform a function that is defined everywhere, to one that is undefined at one point, but we can't do the reverse operation and get back to the original function.

 

[FactChecker]

I'm thinking of the limit of f(x) as x goes to 1. I thought that was the question.

But if you want to graph the original function just put a little circle at f(1) to indicate a discontinuity. I'm not sure what's the point, if we are not talking about the limit.

Just answering this from the original post:

it appears that there is a vertical line in the graph where x=1.

Can anyone explain why this vertical line feature appears in the graph?

Every point on the vertical line is a valid solution of y(x-1)=x²-1. There is no division by 0.

 

[Ned Eterminita]

This actually raises an interesting, or maybe a silly, question. It seems very basic.

Suppose we start with the function f(x) = x + 1. This is defined for all x and is a straight line.

Now we multiply both numerator and denominator by x - 1. So we get the function of the problem, f(x) = (x+1)(x-1)/(x-1) which is undefined for x = 1.

This seems nonsensical. Is it even valid to multiply by (x-1)/(x-1)? If it is, why isn't it equally valid to cancel the x-1 terms?

In other words, it seems strange that we can use algebra to transform a function that is defined everywhere, to one that is undefined at one point, but we can't do the reverse operation and get back to the original function.

I gave my cousin a call, because it seemed strange that he would offer me $20 with no down side.

As it turns out, FactChecker was right and I'm getting that $20 (Thanks FactChecker) but yes he had an ulterior motive. He was using it to butter me up for some crank paper he's drafted called "Indeterminate is not synonymous with undefined" (You'll find it if you google it supposedly) that he wants me to show to my maths lecturer.

Basically, he thinks that 0/0 should be depicted with a vertical line rather than a hole because its indeterminate (aka could be anything) and not undefined like 1/0. It seems that he's reached that conclusion following Aufbauwerk's reasoning. Though maybe entropy is a thing in algebra too?

 

[FactChecker]

I gave my cousin a call, because it seemed strange that he would offer me $20 with no down side.

As it turns out, FactChecker was right and I'm getting that $20 (Thanks FactChecker) but yes he had an ulterior motive. He was using it to butter me up for some crank paper he's drafted called "Indeterminate is not synonymous with undefined" (You'll find it if you google it supposedly) that he wants me to show to my maths lecturer.

Basically, he thinks that 0/0 should be depicted with a vertical line rather than a hole because its indeterminate (aka could be anything) and not undefined like 1/0. It seems that he's reached that conclusion following Aufbauwerk's reasoning. Though maybe entropy is a thing in algebra too?

You should tell your friend that y(x-1)=x²-1 and y=(x²-1)/(x-1) are not the same thing. They do not have the same set of solutions. He will get nowhere with mathematicians.

 

[Aufbauwerk 2045]

This thread reminds me of a bad dream. I feel like I've entered some alternative reality. If f(x) is a function, then there is only one value of f(x) for each value of x that is in the domain of f(x). Vertical lines do not enter into it at all.

Sorry, but all I can think of now is to press the panic button and respond with a video. It's a good video for mental clarity. Good night and goodbye to this thread. I hope someone learned something.

I wonder if they need AI programmers in Belarus.

 

[Mark44]

As it turns out, FactChecker was right and I'm getting that $20 (Thanks FactChecker) but yes he had an ulterior motive. He was using it to butter me up for some crank paper he's drafted called "Indeterminate is not synonymous with undefined" (You'll find it if you google it supposedly) that he wants me to show to my maths lecturer.

Your cousin is correct in saying that the two are not synonymous.

Basically, he thinks that 0/0 should be depicted with a vertical line rather than a hole because its indeterminate (aka could be anything) and not undefined like 1/0. It seems that he's reached that conclusion following Aufbauwerk's reasoning. Though maybe entropy is a thing in algebra too?

$[\frac 0 0]$ is just one of a number of indeterminate forms, which also include $[\frac {\pm \infty} \infty]$, $[\infty - \infty ]$, $[1^\infty]$, and one or two more. They are usually presented in brackets to emphasize that we're not talking about a specific thing, but rather a form. These indeterminate forms usually show up in the context of taking a limit.
The reason they have "indeterminate" in their descriptions is that you can't determine in advance how the limit will turn out. For example, these three limits are all indeterminate forms:
$\lim_{x \to 0} \frac x x$
$\lim_{x \to 0} \frac {x^2} x$
$\lim_{x \to 0} \frac x {x^2}$
Although all are of the ($[\frac 0 0]$) form, the first limit is 1, the second limit is $\infty$, and the third limit is 0.
$\frac 0 0$ on its own is undefined, for the simple reason that division by zero is undefined. If we're working with the limit of a fraction, and the denominator is approaching zero, the limit can exist, even if the numerator is also approaching zero, with the difference being how quickly each of them is approaching zero.

Your cousin's idea that 0/0 should be depicted, let alone as a vertical line -- no mathematician is going to pay any attention to that.

Finally, entropy doesn't enter into the picture at all.

 

[Ned Eterminita]

Your first point is incorrect. If x = 1, the equation you started with becomes y * 0 = 0, which, as already stated, can't be solved for y.

Is this still your view on y(x-1)=x²-1 as in should a hole be depicted?

or are you now in the "y(x-1)=x²-1 and y=(x²-1)/(x-1) not being the same thing" camp?

 

[mfb]

Multiplying by (x-1) leads to additional solutions at x=1. Because you multiply both sides by 0, which makes every equation true. He is adding solutions to the equation (that were not present before) with this step.

 

[Mark44]

Is this still your view on y(x-1)=x²-1 as in should a hole be depicted?

or are you now in the "y(x-1)=x²-1 and y=(x²-1)/(x-1) not being the same thing" camp?

No, the two equations are not equivalent. The graph of the first equation combines two straight lines: the vertical line x = 1 and the line y = x + 1. The second equation is the line y = x + 1 with a "hole" (point discontinuity) at (1, 2).

 

[OmCheeto]

Is there a name for this "Graph what you see without touching ANYTHING!" type maths?
I've never seen it before.

ps. I ran over to wolframalpha and tried to break their maths machine. It just laughed at me, and said; "I can graph that!"

Om; "graph y(x-1)/x(y-1) = (x^2-1)/(y^2-1)"
wa; "

"



I tried doing it by hand, and after about 30 seconds, I decided my brain would explode if I continued.

 

[Ned Eterminita]

No, the two equations are not equivalent. The graph of the first equation combines two straight lines: the vertical line x = 1 and the line y = x + 1. The second equation is the line y = x + 1 with a "hole" (point discontinuity) at (1, 2).

Thanks,

At least now when I explain this in person on Saturday (after collecting my money), I know this is the consensus view on the matter.

Just so I know I have this down pat...

If 3y=6 and y=6/3 are both satisfied by y=2

And 0y=1 and y=1/0 can't be satisfied in either, as there is no value which y can assume to make it true, thus y is Undefined

Then why can 0y=0 be satisfied by any value and thus Indeterminate, whereas y=0/0 can't be satisfied at all and thus Undefined?

Why the incongruity?

Now, would I be right to answer that question with "One cannot divide by zero, regardless of the value of the numerator and the concept of indeterminate only applies when dealing with limits" ?

He's pretty stiff necked, but hopefully i'll get him to see straight.

 

[jbriggs444]

Then why can 0y=0 be satisfied by any value and thus Indeterminate, whereas y=0/0 can't be satisfied at all and thus Undefined?

When you write down a formula, there is an expectation that the formula specifies a unique value. If it does not then the formula is undefined and may not be used. It is not that "y=0/0" cannot be satisfied. It contains an undefined element (0/0). Because of that, it is not a valid equation in the first place.

There is no such problem with "0y=0". It is simply an equation with more than one solution.

 

[Aufbauwerk 2045]

Sorry, it was not a good hint because I was tired.

I will make up for it by giving the correct explanation for you.

We have f(x) = y = (x^2 - 1)/(x-1).

Looks like it has a discontinuity at x = 1, because the denominator = 0.

But the problem goes away when you factor the numerator, so y = f(x) = (x+1)(x-1)/(x-1) = x + 1.

In fact it's just a straight line. f(1) = 2.

[EDIT Sorry, I was a bit careless. I should say lim f(x) as x -> 1 = 2. I've only seen this problem in the context of doing the limit.]

This is my brain on two hours of low quality sleep. But it's hard to get enough sleep. This is why I now work on AI. As long as I write my code when I'm rested, and verify that it's all correct, I never have to worry about it. Human error is the problem, AI is the solution. Computers never get tired or sloppy. HAL 9000, Colossus, and Daystrom's M5 did not make mistakes. It was their programmers or other humans who made the mistakes.

 

[Ned Eterminita]

When you write down a formula, there is an expectation that the formula specifies a unique value. If it does not then the formula is undefined and may not be used. It is not that "y=0/0" cannot be satisfied. It contains an undefined element (0/0). Because of that, it is not a valid equation in the first place.

There is no such problem with "0y=0". It is simply an equation with more than one solution.

So an equation can have more than one solution and formula cannot?

What is the difference between a formula and equation?

 

[jbriggs444]

What is the difference between a formula and equation?

An "equation" has an equal sign and asserts that the entity on the left hand side is identical to the entity on the right. A "formula" or "expression" is a sequence of symbols that specifies a value.

x=y is an equation.
x+y is a formula.

In a typical equation, you put one expression on the left hand side, another expression on the right hand side and judge whether the values of those expressions are identical.

 

[Mark44]

So an equation can have more than one solution and formula cannot?

What is the difference between a formula and equation?

What jbriggs444 said is a little unclear to me. Possibly he was talking about the formula for a function, as opposed to an equation. By definition, a function is a mapping, or pairing, between members of two sets, in which each member of the first set is paired with a single element of the second set. (The two sets don't have to be different.)
For example, for the function f defined as f(x) = 2x + 1, each value of x gives one value of f(x).

On the other hand, an equation can have zero, one, two, or more solutions. The equation $x^2 = 4$ has two solutions. The equation $0x = 0$ has an infinite number of solutions. Neither of these is the formula for a function.

 

[Mark44]

x=y is an equation.
x+y is a formula.

In a typical equation, you put one expression on the left hand side, another expression on the right hand side and judge whether the values of those expressions are identical.

OK, I see what you meant now. What you're calling a formula I usually call an expression.

 

[Ned Eterminita]

The equation $x^2 = 4$ has two solutions. The equation $0x = 0$ has an infinite number of solutions. Neither of these is the formula for a function.

Thanks for clarifying that.

So I want to explain to my cousin why $0x = 0$ and $x = 0/0$ are not the same thing, in that x is undefined in the latter. How would you do it in a simple and concise manner?

 

[FactChecker]

Thanks for clarifying that.

So I want to explain to my cousin why $0x = 0$ and $x = 0/0$ are not the same thing, in that x is undefined in the latter. How would you do it in a simple and concise manner?

Exactly that way. He will have to deal with the fact that division by zero is not defined and get used to it.

 

[jbriggs444]

Thanks for clarifying that.

So I want to explain to my cousin why $0x = 0$ and $x = 0/0$ are not the same thing, in that x is undefined in the latter. How would you do it in a simple and concise manner?

x is not even undefined in the latter. That is because the latter is not even an equation. You cannot have a valid equation which contains an invalid term.

 

[Mr Indeterminate]

Right after my cousin takes my money and puts 1,000s of miles between him and I, does he reveal that he cheated…..

Look, I'm completely willing to accept that 0x=0 and x=0/0 are not the same thing, so long as a reason why this is the case is provided.

I reason the contrary because it's a matter of consistency, consider the below:

If ax=b and y=b/a then x=y

Now the above is correct when a=2 and b=6, as both x and y equal 3. It also correct when b=1 and a=0, as both x and y are undefined. Thus, it makes no sense for it to be incorrect when a=0 and b=0, as it's correct in all other instances.

Ok, now your turn….