## ∞ X 0 = -1?

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### ∞ X 0 = -1?

In my math class (8th grade) we're learning about geometry and slopes, it is my understanding that a perfectly vertical line has a slope of 1/0 (∞ or undefined) and a perfectly horizontal line perpendicular to the vertical line has a slope of 0. We also learned that the product of the slopes of 2 perpendicular lines is always equal to -1. How can this be true, as it would require 0 times infinity to be equal to -1?

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### Re: ∞ X 0 = -1?

You can't multiply by undefined things and expect to get sensible results.

### Re: ∞ X 0 = -1?

Funnily enough I was testing how C++ would handle division by 0 on my machine and other undefined things including what it does with inf*0. On my machine, (1/0)*0=-nan. Nan being 'not a number', so it was being sensible, although 1/0 on it's own returns inf. The interesting part is that it seems to be negative nan, so I came into this topic thinking you'd done similar and concluded that that's where the negative sign came from.

Of course, you shouldn't be doing things with non-numbers like "infinity" and especially "undefined", since you'll always get strange and usually meaningless results. (Just because occasionally computers will let you use them as numbers, doesn't actually make them numbers.)

Of course, you shouldn't be doing things with non-numbers like "infinity" and especially "undefined", since you'll always get strange and usually meaningless results. (Just because occasionally computers will let you use them as numbers, doesn't actually make them numbers.)

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### Re: ∞ X 0 = -1?

The infinity token is used in a variety of ways in math, and in this context it acts as a short form for expressing the slope of a vertical line. However multiplication is only defined for numbers, and infinity is not a number in the usual sense so we can't multiply it. Since vertical lines required a special representation, we should expect that they would require a special rule for testing it's perpendicularity as well. Also note that there does not exist an x such that x*0 = -1 either, so clearly the perpendicular rule doesn't work in that case despite 0 being a perfectly legitimate number.

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### Re: ∞ X 0 = -1?

You're absolutely correct to question what you've been taught there! It cannot be true that a vertical line has a slope of 1/0, because then:

Slope = 1/0

0*Slope = 1

Clearly, that equation cannot hold for any real value of Slope.

Yes, vertical lines are a basic limitation of slope-values (or 'gradients'). 'Infinity' is not a real number and cannot be compared with real numbers; the best I could say is "as the line approaches vertical, the slope approaches infinity".

The problem lies in the assumption that all straight lines can be written as an equation in x and y, in the form 'y = ax + b'. The assumption here is that the line spans all possible x values, which vertical lines do not. A more general equation would be ax + by + c = 0. Then we can get vertical lines by setting b=0 (giving an equation of the form x = d), as well as all other lines still being available.

Footnote: '1/0' can never make any sense. '0/0' has a better chance, since 'x = 0/0' would imply that 0x = 0, i.e. x can take on any value.

Slope = 1/0

0*Slope = 1

Clearly, that equation cannot hold for any real value of Slope.

Yes, vertical lines are a basic limitation of slope-values (or 'gradients'). 'Infinity' is not a real number and cannot be compared with real numbers; the best I could say is "as the line approaches vertical, the slope approaches infinity".

The problem lies in the assumption that all straight lines can be written as an equation in x and y, in the form 'y = ax + b'. The assumption here is that the line spans all possible x values, which vertical lines do not. A more general equation would be ax + by + c = 0. Then we can get vertical lines by setting b=0 (giving an equation of the form x = d), as well as all other lines still being available.

Footnote: '1/0' can never make any sense. '0/0' has a better chance, since 'x = 0/0' would imply that 0x = 0, i.e. x can take on any value.

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### Re: ∞ X 0 = -1?

TLH wrote:A more general equation would be ax + by + c = 0.

At least one of a or b should be nonzero, or else you won't get a line. If you write your lines in this form, then the perpendicular lines would be lines of the form -bx + ay + d = 0. If both a and b are nonzero, then the slopes are -a/b and b/a, respectively, which are negative reciprocals of each other. If only one of a or b is nonzero, then one line will be vertical and the other line horizontal, so one line will have undefined slope.

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### Re: ∞ X 0 = -1?

skeptical scientist wrote:TLH wrote:A more general equation would be ax + by + c = 0.

At least one of a or b should be nonzero, or else you won't get a line. If you write your lines in this form, then the perpendicular lines would be lines of the form -bx + ay + d = 0. If both a and b are nonzero, then the slopes are -a/b and b/a, respectively, which are negative reciprocals of each other. If only one of a or b is nonzero, then one line will be vertical and the other line horizontal, so one line will have undefined slope.

Oh absolutely, setting a=0 and b=0 gives an equation which is satisfied by the entire x-y plane. The point is that the 'slope' is a quantity that describes the orientation of a line in explicit form, thus it shares the limitations of the explicit form that are not present in an implicit form.

### Re: ∞ X 0 = -1?

kikoskia wrote:In my math class (8th grade) we're learning about geometry and slopes, it is my understanding that a perfectly vertical line has a slope of 1/0 (∞ or undefined) and a perfectly horizontal line perpendicular to the vertical line has a slope of 0. We also learned that the product of the slopes of 2 perpendicular lines is always equal to -1. How can this be true, as it would require 0 times infinity to be equal to -1?

Think about this.. can anything actually be equal to infinity? Maybe that will help you understand why this paradox is occurring.

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### Re: ∞ X 0 = -1?

Also, related to division by 0, you say 1/0 = (undefined or infinity). Presumably, you think infinity is what it equals because you've observed that 1/0.1=10, 1/0.01=100, 1/0.000001=1000000, etc. which getting bigger and bigger as the denominator gets closer to 0, and hence 1/0 is intuitively infinity.

However, consider what happens if instead you come at things from the other direction: 1/(-0.1)=-10, 1/(-0.01)=-100, 1/(-0.000001)=-1000000, which is instead getting bigger and bigger in the negative direction, and so that reasoning would lead your intuition towards 1/0 being negative infinity, which couldn't be further from the earlier answer of infinity.

So which is it? Well, it's neither, as it's undefined (and again, infinity isn't actually a number so all this is very hand-wavey anyway). I found those first two arguments most compelling for my intuition when I was first learning that stuff, so hopefully it'll help train your intuition too.

(An interesting case is 0/0 which actually isn't undefined, but rather indeterminate, as depending on the situation it can actually take on any value, as suggested in TLH's post.)

However, consider what happens if instead you come at things from the other direction: 1/(-0.1)=-10, 1/(-0.01)=-100, 1/(-0.000001)=-1000000, which is instead getting bigger and bigger in the negative direction, and so that reasoning would lead your intuition towards 1/0 being negative infinity, which couldn't be further from the earlier answer of infinity.

So which is it? Well, it's neither, as it's undefined (and again, infinity isn't actually a number so all this is very hand-wavey anyway). I found those first two arguments most compelling for my intuition when I was first learning that stuff, so hopefully it'll help train your intuition too.

(An interesting case is 0/0 which actually isn't undefined, but rather indeterminate, as depending on the situation it can actually take on any value, as suggested in TLH's post.)

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### Re: ∞ X 0 = -1?

One nonstandard number system that may be interesting is the real projective line, which is the real numbers plus a single value "at infinity". The system seems to match the idea of the slope of a line reasonably well.

The interesting part is that, in that system, it is true that -1/0 = ∞, and -1/∞ = 0. However, it's not the case that 0*∞ = -1... rather, it is undefined.

The interesting part is that, in that system, it is true that -1/0 = ∞, and -1/∞ = 0. However, it's not the case that 0*∞ = -1... rather, it is undefined.

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### Re: ∞ X 0 = -1?

phlip wrote:One nonstandard number system that may be interesting is the real projective line, which is the real numbers plus a single value "at infinity". The system seems to match the idea of the slope of a line reasonably well.

...which is not at all surprising, given that the "projective" in "real projective line" comes from "projective space", one example of which is the set of lines in the plane which pass through the origin.

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### Re: ∞ X 0 = -1?

Dopefish wrote:(An interesting case is 0/0 which actually isn't undefined, but rather indeterminate, as depending on the situation it can actually take on any value, as suggested in TLH's post.)

Isn't 0/0 undefined as well? We aren't talking about limits here - we're just talking about taking the number 0 and dividing it by the number 0. The idea of this being indeterminate doesn't make sense unless we're talking about which path we took to get to each of those 0s.

double epsilon = -.0000001;

### Re: ∞ X 0 = -1?

Vertical line does not (technically) have a slope. That's probably the easiest way to understand it.

If you look at the standard [imath]\frac{\Delta y}{\Delta x}[/imath] formula for the slope of a line, you will find this is true. Division by zero is undefined, so the slope is undefined. This is the reason why the "slope rule" fails: there is no slope!

The only sense in which we can say there's a slope of [imath]\infty[/imath] is as an exemption to the rule, so to speak. I'm sure you understand how that particular limit comes about. But an even better way to personally think about it, to avoid confusion, is to think "infinite slope" instead of "slope of [imath]\infty[/imath]"; the latter suggests [imath]\infty[/imath] might be a number, when in fact it's not. It's the notion that slopes that get "more vertical" get higher and higher without bound.

In fact - though your teachers may not explain this explicitly - each notion of [imath]\infty[/imath] (at least in middle and high school) is just a way of expressing something that gets bigger and bigger without bound, and it's helpful to try to think of what each thing might be.

If you look at the standard [imath]\frac{\Delta y}{\Delta x}[/imath] formula for the slope of a line, you will find this is true. Division by zero is undefined, so the slope is undefined. This is the reason why the "slope rule" fails: there is no slope!

The only sense in which we can say there's a slope of [imath]\infty[/imath] is as an exemption to the rule, so to speak. I'm sure you understand how that particular limit comes about. But an even better way to personally think about it, to avoid confusion, is to think "infinite slope" instead of "slope of [imath]\infty[/imath]"; the latter suggests [imath]\infty[/imath] might be a number, when in fact it's not. It's the notion that slopes that get "more vertical" get higher and higher without bound.

In fact - though your teachers may not explain this explicitly - each notion of [imath]\infty[/imath] (at least in middle and high school) is just a way of expressing something that gets bigger and bigger without bound, and it's helpful to try to think of what each thing might be.

### Re: ∞ X 0 = -1?

Dason wrote:Dopefish wrote:(An interesting case is 0/0 which actually isn't undefined, but rather indeterminate, as depending on the situation it can actually take on any value, as suggested in TLH's post.)

Isn't 0/0 undefined as well? We aren't talking about limits here - we're just talking about taking the number 0 and dividing it by the number 0. The idea of this being indeterminate doesn't make sense unless we're talking about which path we took to get to each of those 0s.

I might be able to accept indeterminate things as being a subset of undefined, although I've always considered it to be one or the other (or defined). I'm not sure it's a relevent distinction though, since either way you're in trouble and need to go about things in away that avoids division by 0.

However I do think indeterminate is the appropriate term for 0/0 no matter how you come about it, since if you consider a possible value for it to evaluate to 'C', you have 0/0=C, or 0=C*0, which is true for every value of C and you just can't determine which one is the 'right' one. Under certain conditions, sure you can do some limit taking to determine what value of C is the right one for a specific instance, but the 0/0 entity would be indeterminate regardless of whether tricks exist to go ahead and determine it.

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### Re: ∞ X 0 = -1?

Indeterminate forms aren't directly related to undefined values... you can have undefined things that are determinate (like 1/0) and you can have indeterminate things that are still defined (like 0

^{0}). I went into more detail a while back.Code: Select all

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### Re: ∞ X 0 = -1?

Dopefish wrote:Also, related to division by 0, you say 1/0 = (undefined or infinity). Presumably, you think infinity is what it equals because you've observed that 1/0.1=10, 1/0.01=100, 1/0.000001=1000000, etc. which getting bigger and bigger as the denominator gets closer to 0, and hence 1/0 is intuitively infinity.

However, consider what happens if instead you come at things from the other direction: 1/(-0.1)=-10, 1/(-0.01)=-100, 1/(-0.000001)=-1000000, which is instead getting bigger and bigger in the negative direction, and so that reasoning would lead your intuition towards 1/0 being negative infinity, which couldn't be further from the earlier answer of infinity.

So which is it? Well, it's neither, as it's undefined (and again, infinity isn't actually a number so all this is very hand-wavey anyway). I found those first two arguments most compelling for my intuition when I was first learning that stuff, so hopefully it'll help train your intuition too.

(An interesting case is 0/0 which actually isn't undefined, but rather indeterminate, as depending on the situation it can actually take on any value, as suggested in TLH's post.)

A vertical line being both positive and negative at the same time actually makes perfect sense (some how) because it seems like its in between pointing "up" (positive)and "down" (negative)

### Re: ∞ X 0 = -1?

TLH wrote:Footnote: '1/0' can never make any sense. '0/0' has a better chance, since 'x = 0/0' would imply that 0x = 0, i.e. x can take on any value.

Ah, but you've multiplied both sides of the equation by zero in order to get the statement "'x = 0/0' would imply that x=0/0", i.e.:

0*x=0*(0/0)

0*x=0*1

0*x=0

Which is, of course, improper and commonly used to prove that 1 = 2. However, it is true that 0x=0 is valid for all x.

A much simpler way to deal with dividing by zero is to fall back to the old definition of division, which is repeated subtraction. In other words, the question "What is A divided by B?" becomes the question "How many times can I grab B out of A until there is nothing left of A"

For example, 20/5 is 20-> (1) 15 -> (2) 10 -> (3) 5 -> (4) 0. Answer: 4.

How many times can you grab 7 dinars out of a pile of 42 dinars until there are no dinars left? 6 times. Now, what is 42/0? How many times can you grab 0 dinars out of the pile until there are no dinars left? You could do infinite zero-grabs, but you'll still have 42 dinars left, so 42/0 is not equal to infinity. Since no amount of zero-grabbing is going to get the pile to be gone, the "value" of 42/0 is "undefined". Of course, even the repeated subtraction breaks down in the case of 0/0, because then the question is "How many times can you grab 0 dinars out of a pile that is already empty until there are no dinars left?" The answer is "However many times you want", which mathematicians refer to as "indeterminate".

As for the original poster, you can either have a rule that vertical lines have no defined slope, or just go with slope = pi / 2.

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### Re: infty X 0 = -1?

There's nothing wrong with multiplying both sides of an equation by 0. It's not exactly helpful ever, and it will introduce solutions, but it never leads you from a statement which is true to a statement which is false. Which means it's consistent.

This is unlike dividing both sides by zero, which can lead you from a statement which is true to a statement which is false.

Also, a slope of pi/2 is this, not a vertical line. You're confusing gradients with angles.

This is unlike dividing both sides by zero, which can lead you from a statement which is true to a statement which is false.

Also, a slope of pi/2 is this, not a vertical line. You're confusing gradients with angles.

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### Re: ∞ X 0 = -1?

I think DavidRoss is suggesting using arctan(dy/dx) as a measure of gradient instead of using dy/dx.

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### Re: infty X 0 = -1?

phlip wrote:There's nothing wrong with multiplying both sides of an equation by 0. It's not exactly helpful ever, and it will introduce solutions, but it never leads you from a statement which is true to a statement which is false. Which means it's consistent.

This is unlike dividing both sides by zero, which can lead you from a statement which is true to a statement which is false.

Also, a slope of pi/2 is this, not a vertical line. You're confusing gradients with angles.

Sheesh, was I sloppy. Multiplying both sides by zero is indeed not the problem. The problem is cancelling out the (0/0) term.

And, I was not confused, but sloppy in that I meant angle = pi/2. As a programmer, lots of stuff makes me nervous, like having to code for a special case, and have someone later come along and clean up my code, tossing out the dx=0 handler and have the program crash in undebuggable ways. Instead, I would ditch gradients/slopes altogether and use angle to represent the orientation of a line. And as the input, not as the output of doing an arctan(dy/dx) calculation on the input.

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