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Here is given $A(x_1,y_1), B(x_1,y_2), C(x_2,y_3)$ and $D(x_3,y_3)$. I have recently read that, multiplication of two perpendicular lines is always $-1$.

From the above graph, the slope of $AB, m_1 = \frac{y_2-y_1}{x_1-x_1} = \frac{y_2-y_1}{0} = \infty$

and the slope of $CD, m_2 = \frac{y_3-y_3}{x_3-x_2} = \frac{0}{x_3-x_2} = 0$

And then we get $m_1m_2 = -1$

$\infty*0 = -1$

Here it is clearified that multiplication of infinity and zero always leads to $-1$ from the above formula. Then in which fact, do we strongly put emphasis on? $\infty*0 = -1$ or $\infty = \frac{-1}{0}$

If it indicates the 1st condition, then why $\infty*0$ would be only equal to $-1$. It could have been resulted in any real number. And how we define the infinity in this case?

Any kind of conception would be greatly helpful for me to remove my ignorance. Thanks in advance.

Anirban Niloy
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    As you've no doubt guessed, this case is an edge case that doesn't really follow the rule $m_1 m_2 = -1$ exactly because multiplying infinity and zero is extremely poorly defined. – Cameron Williams Mar 03 '19 at 16:53
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    A vertical line has no slope (at least not in the usual sense). Besides the usual issues with $\infty$ as a number, an issue is that we could also $-\infty$ consider to be the "slope" because it is not clear whether the vertical line is "strictly decreasing" or "strictly increasing". – Peter Mar 03 '19 at 16:54
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    @CameronWilliams I prefer Haran's answer to this approach. You are technically correct, but his approach shows the intuition that in some sense, had it been defined, this is what it should have been... – gt6989b Mar 03 '19 at 16:58
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    The rule that if two lines in $\mathbb R^2$ are orthogonal then the product of their slopes is $-1$ only holds if neither line is vertical. I think the source you're reading should have omitted this claim that $\infty \times 0 = -1$. We could define $\infty \times 0$ to be $-1$, but this isn't necessary and doesn't seem very helpful. (In some other contexts, like measure theory, the convention that $\infty \times 0 = 0$ is useful.) – littleO Mar 03 '19 at 16:59

3 Answers3

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Good question! The trick is: infinity is not a number (in most frameworks).

When we write ∞, we're using it as a convenient shorthand for "what happens when this variable gets arbitrarily large". And if you try to use it as a number, things go horribly wrong.

When you ask about something like $0 \times \infty$, what you're usually asking is: "what happens to $0 \times k$, as $k$ gets arbitrarily large?" Using limit notation, we'd say "what is $\lim_{k \rightarrow \infty} 0 \times k$?"

And sometimes this can be answered: we can see that, no matter how big $k$ gets, $0 \times k$ is always zero. So $\lim_{k \rightarrow \infty} 0 \times k = 0$. Sometimes it can't. Suppose we instead asked about $\frac{1}{k} \times k^2$: the first part is going to zero, the second part is going to infinity. But this time, as $k$ gets bigger and bigger, the product just keeps getting bigger and bigger too. So in this case, our "$0 \times \infty$" ends up being infinite.

So the answer to $0 \times \infty$ is: "it depends how you get there". Infinity isn't a number, and it doesn't act like one: rather, it's a shorthand for "let's see what happens when we let this variable get arbitrarily large".

Draconis
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  • Superb answer indeed (+1) !! – Peter Mar 03 '19 at 17:08
  • I don't know if this would help @AnirbanNiloy or just be confusing, but the limit of $0\times\infty$ could be $-1$ if we start with something like other perpendicular lines of slope $2$ and $-1/2$, and replace the $2$ with larger and larger numbers to get close to the picture of the vertical and horizontal line. – Mark S. Mar 03 '19 at 20:55
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The answer by Haran is incorrect, and the answer by Draconis is a bit unclear. Specifically, one cannot say that "$0·∞$ can take any value". In the first place, "$1/0$" is ill-defined when $1,0$ are ordinary reals, so there is no way to get any such thing as a real $\infty$. Thus "$0·∞$" is simply a meaningless expression unless you give a rigorous definition of "$∞$" as some mathematical object. For example, you could very well let $∞ = 8$ even if it is unconventional.

Of course, in real analysis we often have limits to infinity, but that is not the same as just doing an illegal division by zero. We normally define "$\lim_{x→c} \cdots = ∞$" as merely notational short-hand for some formal statement, and not that the limit expression has a value that is equal to something called $∞$. That is why we cannot write "$\lim_{x→0} \cos(1/x) ≠ ∞$" because in the first place "$\lim_{x→0} \cos(1/x)$" is undefined.

Also, although it is true that $\lim_{x→0} 1/x^2 = ∞$, it is meaningless to write "$1/0^2 = ∞$" (unless you're not working with ordinary reals, in which case you must know precisely what you're doing).

Draconis' answer is unclear, because you must very clearly understand that "$0·∞$" is not an expression with a value. Rather, it is merely a string of symbols. (See this related post about the difference between $0^0$ and the form "$0^0$".)

As for your initial question, your mistake is right at the beginning; your equation for a vertical line is wrong. In general, the equation of a line in the $x,y$-plane is { $ax+by=c$ } for some reals $a,b,c$. You cannot divide the equation by $b$ if $b=0$. So you cannot even get to the point of 'computing' the slope of a vertical line.

But there are correct versions of the facts you're interested in:

Given any line $L$ in the $x,y$-plane with equation { $ax+by=c$ }, the slope of $L$ is the ratio $-a:b$.

Given any perpendicular lines $L,M$, their slopes $a:b$ and $c:d$ satisfy $a:b = -d:c$.

Note that $1:0$ and $0:1$ are both valid ratios! "$0:0$" is not a valid ratio. (At least one must be nonzero.) Two ratios $a:b$ and $c:d$ are equal iff $ad = bc$. In particular, notice that $1:0 = -1:0$.

Also note that this formulation handles all kinds of lines correctly (without facing illegal division by zero).

Community
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user21820
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  • I noticed my mistake of constructing equation for a vertical line. Does it mean that $\frac{\delta y}{\delta x}$ isn't valid for slope of any vertical line parallel to $y$ axis? – Anirban Niloy Mar 26 '19 at 16:33
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    @AnirbanNiloy: Right. Intuitively, it's because there's no change in $x$ as you move along the line, so you can't possibly have "(change in $y$) / (change in $x$)". I can make it more precise if you want, but basically if we fix a point on the curve and move another point along the curve towards it, we need $Δx→0$ (eventually stays close enough to zero but does not reach zero), otherwise "$\frac{Δy}{Δx}$" is not even well-defined. Only after we have that condition satisfied, then we can see whether $\frac{Δy}{Δx}$ tends to a fixed real $r$, in which case we define $\frac{dy}{dx} = r$. – user21820 Mar 26 '19 at 17:30
  • I understand what you're doing with "ratios" like $1:0$ because I'm familiar with $\mathbb{R}\mathbf{P}^1$, but do you have a source uses the word "ratio" and would support "$1:0$ [is] a valid ratio"? It seems okay in the context of "odds" as in a table on wikipedia, but outside of that context I would think most sources would not allow a ratio of $1:0$. – Mark S. Mar 27 '19 at 01:17
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    @MarkS.: I don't have a source at my fingertips, because I learnt this when I was young as part of olympiad preparations. When I learnt it, the distinction is that a ratio is simply a pair of reals modulo the equivalence I gave (which as you observed is one way to construct the real projective line) whereas a quotient $a/b$ is only defined when $b≠0$. Also, ratios extend naturally to longer tuples, which are used for barycentric coordinates (as the Wikipedia article on ratios also says). Yes, many places define ratios via quotients, but that's not what "ratio" means to me in English. – user21820 Mar 27 '19 at 04:16
  • There's of course the possibility that the source I learnt from was influenced by projective geometry. =) – user21820 Mar 27 '19 at 04:16
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When you take the limit for your answer at these expressions, it happens that the limit tends to $-1$. There is nothing special about the value $-1$ itself. Consider $n \cdot \frac{2}{n}$ where $n$ tends to infinity. The value is always $2$. However, this does not mean you can directly set $n= \infty$ for a proper answer. Expressions like $0 \cdot \infty$ are not undefined expressions which have no answer, but instead, indeterminate forms, which can take any answer (limit) based on the situation.

Haran
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    Correction: $0 \cdot \infty$ is an indeterminate form, but it is undefined as an expression. Compare $0^0$, which is also an indeterminate form, but as an expression is defined as $0^0 = 1$. – eyeballfrog Mar 03 '19 at 16:58
  • @eyeballfrog: say what?? $0^0$ is undefined in my book (even "as an expression"). – TonyK Mar 03 '19 at 16:59
  • @TonyK It's commonly defined to be 1 to make Taylor series work out more elegantly. – Draconis Mar 03 '19 at 17:03
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    $0^0$ is soemthing different. I wonder why so many people disagree the definition $0^0=1$ , which is used in some math-softwares and allows to generalize many useful theorems. Not sure, whether $0^0=0$ has any advantage. What finally convinced me was the fact $$\lim_{x\rightarrow 0+0} x^x=1$$ – Peter Mar 03 '19 at 17:04
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    Comments on $0^0$ are off-topic here. You may read many posts on the question already in math.se . – GEdgar Mar 03 '19 at 19:40
  • Sorry, your answer is wrong; one cannot say that "$0·∞$" is defined, nor that it can take any answer... Please read the linked post in my answer to understand the difference between an expression and its value (if any). – user21820 Mar 27 '19 at 04:31
  • @Peter: You may also want to read that linked post because the value of a limit (if it exists) has no relation to its form, just as the limit of a sequence of positive reals may not be positive. The best argument for $0^0 = 1$ is that it's the base case for the recursive definition of exponentiation by a natural (see here), which is why it arises in almost every theorem with natural exponentiation, such as the binomial theorem. Same in other rings, we define $f^0 = id_S$ for any function $f$ on $S$, and $A^0 = I_n$ for any $n×n$ square matrix. – user21820 Mar 27 '19 at 04:37