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### Infinity Solution

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Date: 04/12/2001 at 03:17:14
From: Mr. Chen
Subject: Infinity puzzle

I was reading a maths text on my way home today when I came across
this section on quadratic equations:

If you are to solve a quadratic equation in which the x^2 terms
cancel out, you must include the solution (x = infinity) where
necessary.  E.g.  Solve x^2 - 3x = x^2 + 1. Then in this case you
obtain the answer x = -1/3 and x = infinity.

So I was thinking: if this is really the case, what about the
equation 1 + 2x = 3 + 2x?  Technically speaking, the lines of the two
equations don't intersect at all, so there shouldn't be any solution.
But if we consider the question again, it appears that x = infinity is
the solution. Of course a solution that is infinite is meaningless,
but let's say x is replaced by tan x. The equation tan x = infinity
can be solved, and gives the answer x = pi/2.

So I extended this to reason that since this is possible, what about
other equations? I tried considering the quadratic equation in which
the x^2 terms don't cancel out. Let's say, x^2 + 2x = 3x^2 - x. The
solution is obviously x = 0 and x = 3/2. But x = infinity can be the
solution too, right? I mean, if you consider the graphs of the two
functions on both sides of the equation, i.e. f(x) = x^2 + 2x and
g(x)= 3x^2 - x, both f(x) and g(x) will tend to infinity when x tends
to infinity as well. So technically speaking, does it mean that
infinity is the universal solution of any equation??

I thought about it, but it appeared that such a speculation would
bring about some problems.  For example, let's say I'm to solve this
quadratic equation: x^2 + 3x + 10 = 0.  This equation has imaginary
roots. However, if you adjust the equation a little to become
x^2 = -3x - 10, the solution would be the point(s) of intersection
between the graphs of the two functions f(x) = x^2 and g(x) = -3x - 10
but if you carried out the argument I did earlier (that both functions
tend to infinity as x tends to infinity as well), you supposedly get
the solution x = infinity. But looking back at the original equation,
x^2 + 3x + 10 = 0, and substituting the value of x = infinity, I
realised that I get the equation infinity = 0. What is going on?

I'm sure there's a flaw in my logic somewhere, but I can't seem to
figure it out.
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Date: 04/12/2001 at 12:36:51
From: Doctor Peterson
Subject: Re: Infinity puzzle

Hi, Mr. Chen.

I don't agree with the text's statement in the first place: we don't
consider infinity to be a solution to any equation, because it is not
a valid number. For some purposes you can think of infinity as a
solution (in the sense that, as the variable becomes larger, the
equation comes closer to being true), but that is not a normal way to
talk about solutions.

But even if we accept this concept, you have to be careful about what
it means. I think the claim that infinity is a solution of the given
quadratic is based on the fact that, for large x, the squared terms
will overwhelm the other terms, so that both sides are essentially
equal. You can see that by dividing all terms by x^2:

x^2 - 3x = x^2 + 1

1 - 3/x = 1 + 1/x^2

As x becomes large, this approaches 1 = 1, and becomes more nearly
true.

But the equation as given is really the same as

-3x = 1

and this does NOT become true when x becomes infinite. The left side
becomes infinite, and the right side does not. This is just one hint
that working with infinity is not "safe"; things don't behave right.
What's happening is that if x is infinite, we are subtracting infinity
from both sides, and infinity minus infinity is indeterminate. Look at
it one way and it's zero, another way and it's infinite.

Now, take your next equation:

1 + 2x = 3 + 2x

Here, as in the quadratic, the coefficients of the highest terms are
the same, so when we divide by x we get an equation that remains
finite at infinity, and becomes "true":

1/x + 2 = 3/x + 2

The fact that the lines are parallel, and therefore meet at the "point
at infinity" (in some geometries) agrees with this.

Now look at

x^2 + 2x = 3x^2 - x

Because the coefficients of x^2 are NOT the same, this does not
become "true" when x approaches infinity:

1 + 2/x = 3 - 1/x

becomes 1 = 3 when x is infinite. If you graph the two sides, the
curves do not come close when x is large; one remains about 3 times
the other. The two real solutions you get are the only solutions even
if you include infinity.

So your speculation that infinity is a solution of all equations is,
as you found, nonsense. The whole claim that you can consider infinity
to be a solution of a quadratic is highly questionable, and they
should at least have (a) defined exactly what they mean, and (b) told
you what they mean by "where necessary." I think they meant "when the
coefficients of the highest order terms are the same, and infinity
makes sense in the problem you are solving"; and even then I'm not
sure this really makes sense. I'm very curious to know in what context
this claim was made; perhaps it makes sense in connection with some
particular type of problem, where it is not really the quadratic
equation itself they are trying to solve.

You'll want to look at our FAQ on dividing by zero, which raises some
of the questions you have to consider if you try to treat infinity as
a number:

http://mathforum.org/dr.math/faq/faq.divideby0.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Number Theory

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