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Infinity is not a number

Infinity and strange math · level 5, lesson 3 of 6

A question to start

Hilbert's hotel has infinitely many rooms, numbered 1, 2, 3, ..., and every room is full. Now some guests check out. Try three different mornings:

  • Morning A: every guest checks out. Rooms still full: none, 0.
  • Morning B: the guests in the even rooms 2, 4, 6, ... check out. Rooms still full: 1, 3, 5, ..., infinitely many.
  • Morning C: every guest in a room numbered 4 or more checks out. Rooms still full: 1, 2 and 3, so 3.

Each morning, infinitely many guests were there and infinitely many left. So each morning is "infinity minus infinity". The answers were 0, infinity and 3. Which one is the real value of \(\infty - \infty\)?

None of them. This lesson explains why \(\infty\) does not behave like a number, why some calculations with it have no single answer, and why the same thing goes wrong when you divide by zero.

The idea

What the symbol ∞ really means

In lesson 1 we wrote "\(n \to \infty\)" for "\(n\) grows without end". The symbol \(\infty\) there is shorthand for a process, not a place you arrive at. When we say a sequence tends to infinity, we mean:

Name any number, however huge. From some position on, every term is bigger than it.

For example \(a_n = n^2\) tends to infinity: to beat a million, go past term 1000 (since \(1000^2 = 1000000\)). This is a closeness game again, with "bigger than any number you name" in place of "closer than any amount you name".

Notice that \(\infty\) never appears as an actual value. Every term of every sequence is an ordinary number. That is the key.

If ∞ were a number, 1 would equal 0

Suppose, just to see what happens, that \(\infty\) were an ordinary number that follows the usual rules of arithmetic. In Hilbert's hotel, one new guest arrives and everyone moves up a room; the hotel is still exactly as full. So we would want

\[\infty + 1 = \infty.\]

Now use an ordinary rule: you can subtract the same number from both sides of an equation. Subtract \(\infty\) from both sides, and you get \(1 = 0\).

That is nonsense, so the supposition was wrong: \(\infty\) cannot be a number that obeys the usual rules. This is an example of proof by contradiction: assume something, follow the rules, reach something impossible, and conclude the assumption was false (more in Logic).

∞ − ∞ has no single answer

Mathematicians give \(\infty - \infty\) a meaning through sequences. Take two sequences that both tend to infinity, and look at what their difference does.

First sequence \(a_n\) Second sequence \(b_n\) Difference \(a_n - b_n\) The difference tends to
\(n\) \(n\) \(0\) 0
\(n + 5\) \(n\) \(5\) 5
\(n^2\) \(n\) \(n^2 - n = n(n - 1)\) infinity
\(n\) \(n^2\) \(n - n^2\) minus infinity
\(n + (-1)^n\) \(n\) \((-1)^n\) nothing (no limit)

Every row is "infinity minus infinity", and the answers are all different. The answer depends on how fast each sequence grows, which the symbol \(\infty\) forgets. So \(\infty - \infty\) is called an indeterminate form: "indeterminate" means "not settled by the information given". It is not that the answer is unknown; there is no single answer to know.

The hotel mornings are the same thing in picture form: which guests leave decides how many stay.

∞ ÷ ∞ and 0 × ∞: also no single answer

The same thing happens with division.

\(a_n\) \(b_n\) \(\dfrac{a_n}{b_n}\) Tends to
\(2n\) \(n\) \(2\) 2
\(n^2\) \(n\) \(n\) infinity
\(n\) \(n^2\) \(\dfrac1n\) 0
\(3n + 1\) \(n\) \(3 + \dfrac1n\) 3

And with "something shrinking to 0, times something growing to infinity":

Shrinks to 0 Grows to infinity Product Tends to
\(\dfrac1n\) \(n\) \(1\) 1
\(\dfrac5n\) \(n\) \(5\) 5
\(\dfrac1n\) \(n^2\) \(n\) infinity
\(\dfrac{1}{n^2}\) \(n\) \(\dfrac1n\) 0

So \(\dfrac{\infty}{\infty}\) and \(0 \times \infty\) are indeterminate forms too. A shrinking number and a growing number are in a tug of war, and either side can win, or they can draw at any value.

Some combinations are safe. If \(a_n\) and \(b_n\) both tend to infinity, then \(a_n + b_n\) and \(a_n \times b_n\) do too (two huge positive numbers added or multiplied are even huger). And \(a_n + 5\) tends to infinity. People sometimes write these as "\(\infty + \infty = \infty\)", but always remember it is shorthand for a fact about sequences.

Why we never divide by zero

Zero causes a similar problem, and the reason is simple once you remember what division means.

"\(12 \div 3 = 4\)" means: 4 is the number that, multiplied by 3, gives 12. Division undoes multiplication:

\[12 \div 3 = 4 \quad\text{because}\quad 4 \times 3 = 12.\]

Now try \(6 \div 0\). We need a number that, multiplied by 0, gives 6. But anything times 0 is 0, never 6. There is no such number. So \(6 \div 0\) has no answer.

Try \(0 \div 0\). We need a number that, multiplied by 0, gives 0. Now every number works: \(5 \times 0 = 0\), \(-17 \times 0 = 0\), \(\frac12 \times 0 = 0\). So \(0 \div 0\) has every answer, which means it has no single answer.

Either way, division by zero cannot be given a value that keeps arithmetic working. We say it is undefined.

Could we say 6 ÷ 0 = ∞?

It is tempting. Divide 6 by smaller and smaller numbers:

\(x\) 1 0.1 0.01 0.001 \(-0.001\) \(-0.01\)
\(6 \div x\) 6 60 600 6000 \(-6000\) \(-600\)

As \(x\) shrinks towards 0 from the positive side, \(\dfrac6x\) grows towards infinity. But from the negative side it plunges towards minus infinity. So "the" value at 0 would have to be both \(+\infty\) and \(-\infty\) at once, and we already know \(\infty\) is not a number anyway. The graph of this kind of function is the subject of lesson 4.

0 ÷ 0 is where calculus lives

Here is the surprise: the forms with no single answer are not useless. They are where the interesting mathematics happens. As \(x\) shrinks to 0 (but never equals 0):

  • \(\dfrac{x}{x} = 1\) for every such \(x\), so it tends to 1.
  • \(\dfrac{2x}{x} = 2\), so it tends to 2.
  • \(\dfrac{x^2}{x} = x\), so it tends to 0.

All three are "\(0 \div 0\)" in the limit, and each has a perfectly good limit of its own. Calculus finds the speed of a falling stone at one instant this way: distance gone divided by time taken, with both shrinking to 0. See Speed at an instant in Change.

And one famous example. \(\left(1 + \frac1n\right)^n\) is "a number close to 1, raised to a huge power", a form written \(1^\infty\). You might guess the answer is 1, since 1 to any power is 1. But:

\(n\) 1 2 10 100 1000 1,000,000
\(\left(1 + \frac1n\right)^n\) 2 2.25 2.5937... 2.7048... 2.7169... 2.71828...

The sequence settles at a special number, about 2.71828, called \(e\). It turns up all over mathematics, from compound interest to radioactive decay. \(1^\infty\) is indeterminate too.

Worked examples

1. Find the hidden division by zero. Here is a "proof" that \(2 = 1\).

  1. Let \(a = b\), with \(a\) not zero.
  2. Multiply both sides by \(a\): \(a^2 = ab\).
  3. Subtract \(b^2\) from both sides: \(a^2 - b^2 = ab - b^2\).
  4. Factor both sides: \((a + b)(a - b) = b(a - b)\).
  5. Divide both sides by \((a - b)\): \(a + b = b\).
  6. Since \(a = b\): \(2b = b\), so \(2 = 1\).

The broken step is 5. Since \(a = b\), the bracket \(a - b\) equals 0, so step 5 divides by zero. Steps 1 to 4 are all true (step 4 says \(0 = 0\) in disguise), but step 5 is not allowed. This is why "never divide by zero" is a rule and not just a habit.

2. Infinity minus infinity in the hotel. The hotel is full. The guests in rooms 11, 12, 13, ... check out. How many guests remain?

Rooms 1 to 10 are still occupied, so 10 guests remain. Infinitely many left, from infinitely many, and 10 are left: one more example showing \(\infty - \infty\) has no fixed value.

3. A tug of war. Find the limit of \(\dfrac{n^2 + 3n}{n^2}\).

  1. Top and bottom both tend to infinity, so this is the form \(\dfrac{\infty}{\infty}\). That tells us nothing yet.
  2. Split it: \(\dfrac{n^2 + 3n}{n^2} = \dfrac{n^2}{n^2} + \dfrac{3n}{n^2} = 1 + \dfrac3n\).
  3. \(\dfrac3n\) tends to 0, so the limit is 1.

4. Shrinking times growing. Find the limit of \(\dfrac{1}{n} \times (2n + 7)\).

The first factor shrinks to 0 and the second grows to infinity: the form \(0 \times \infty\). Multiply out: \(\dfrac{2n + 7}{n} = 2 + \dfrac7n\), which tends to 2.

Try it

  1. Explain, using multiplication, why \(20 \div 5 = 4\), and why \(20 \div 0\) has no answer.
  2. Find two sequences, both tending to infinity, whose difference tends to 10.
  3. Find the limit, if it exists: (a) \((n + 7) - (n + 2)\); (b) \(\dfrac{4n}{2n}\); (c) \(\dfrac{n^2}{n^3}\); (d) \(\dfrac{n^3}{n^2}\).
  4. Hilbert's hotel is full. (a) The guests in rooms numbered more than 100 check out. How many remain? (b) The guests in the odd rooms check out. How many remain?
  5. Find the mistake: "\(0 \times 1 = 0\) and \(0 \times 2 = 0\), so \(0 \times 1 = 0 \times 2\). Dividing both sides by 0, \(1 = 2\)."
  6. Work out \(\dfrac{6}{x}\) for \(x = 0.01\), \(x = -0.01\) and \(x = 0.0001\). What happens as \(x\) gets close to 0?
  7. Find the limit of \(\dfrac{5n - 2}{n}\) and of \(\dfrac1n \times n^3\).
  8. Work out \(\left(1 + \frac1n\right)^n\) for \(n = 1, 2, 3\) as exact fractions. Explain in your own words why the form \(1^\infty\) does not have to give 1.
Answers
  1. \(20 \div 5 = 4\) because \(4 \times 5 = 20\). For \(20 \div 0\) we would need a number that gives 20 when multiplied by 0, but every number times 0 is 0. No number works, so there is no answer.
  2. For example \(a_n = n + 10\) and \(b_n = n\). Both grow without end, and \(a_n - b_n = 10\) for every \(n\), so the difference tends to 10. (Many other pairs work, such as \(n^2 + 10\) and \(n^2\).)
  3. (a) The difference is always 5, so the limit is 5. (b) \(\dfrac{4n}{2n} = 2\): limit 2. (c) \(\dfrac{n^2}{n^3} = \dfrac1n\): limit 0. (d) \(\dfrac{n^3}{n^2} = n\): tends to infinity, no limit.
  4. (a) Rooms 1 to 100 stay full: 100 guests. (b) The even rooms stay full: infinitely many guests.
  5. Dividing by 0 is not allowed. The true statement \(0 \times 1 = 0 \times 2\) says nothing about 1 and 2; it is just \(0 = 0\).
  6. \(\dfrac{6}{0.01} = 600\), \(\dfrac{6}{-0.01} = -600\), \(\dfrac{6}{0.0001} = 60000\). From the positive side the values grow without end; from the negative side they become hugely negative. There is no single value they head for.
  7. \(\dfrac{5n - 2}{n} = 5 - \dfrac2n\), which tends to 5. And \(\dfrac1n \times n^3 = n^2\), which tends to infinity (a \(0 \times \infty\) form where infinity wins).
  8. \(n = 1\): \(2^1 = 2\). \(n = 2\): \(\left(\frac32\right)^2 = \frac94 = 2.25\). \(n = 3\): \(\left(\frac43\right)^3 = \frac{64}{27} \approx 2.370\). The base is a bit more than 1, and the power grows. A bit more than 1 makes the answer grow, a huge power makes it grow a lot, but the "bit more" shrinks at the same time. The two effects pull against each other; here they settle at \(e \approx 2.718\), not at 1.

Common mistake

Treating ∞ like a big number. Writing \(\infty - \infty = 0\) or \(\dfrac{\infty}{\infty} = 1\) looks like ordinary arithmetic, but it is wrong: the tables above show these forms can come out as anything. Whenever you see one, go back to the sequences and work out what they really do.

"6 ÷ 0 = 0" or "6 ÷ 0 = 6". A calculator says "error" for a reason. Check any answer by multiplying back: \(0 \times 0 = 0\), not 6, and \(6 \times 0 = 0\), not 6. No number passes the check.

Where it leads