### The Product of Two Natural Numbers

YAP Von Bing

February 14, 2022

Let \(a\) and \(b\) be natural numbers, i.e., 1, 2, 3, etc. Let their product be \(n\), so that \(n = a \times b\), or more briefly \(n = ab\). You might agree that \(n\) cannot be smaller than \(a\) or \(b\), i.e., it must be the case that \(a \le n\) and \(b \le n\). This is easy enough to see in numerous examples. \(2 \times 3 = 6\) , and 6 is greater than both 2 and 3. \(1 \times 5 = 5\), and both 1 and 5 are less than or equal to the product 5. The second example shows that the inequality may not strict, i.e., \(a\) or \(b\) might be equal to \(n\).

How can we demonstrate the general fact, or {\bf theorem}?

How can we prove that for any natural numbers \(a\) and \(b\) with product \(n = ab\), both \(a \le n\) and \(b \le n\) hold? We cannot check every case, even with the help of the most powerful computers, because there are infinitely many natural numbers.

Here is a visual approach, which you should try with pen and paper. \(n\) identical squares can be arranged in a rectangle with \(a\) rows, each having \(b\) squares. Clearly, the total \(n\) cannot be less than the number of squares in a column, \(a\), or the number of squares in a row, \(b\). We may translate the pictorial argument into algebra.

\( n = ab = a + a(b-1) \ge a \)

The second equality is key, which reflects the separation of a column from the others. Since \(b-1 \ge 0\), \(a(b-1) \ge 0\), hence we are certain that \(n \ge a\). What has just happened deserves special attention. Inspired by the picture, the algebraic work has resulted in a {\bf general argument}, which works for all the infinitely many cases.

Incidentally, both the picture and algebra also show that in the special case \(b > 1\), actually \(n > a\). You might want to make a similar argument for why \(n \ge b\).

Let us suppose that \(a \le b\). Can you say something stronger than \(a \le n\)? For instance, can you find an expression \(f(n)\) in terms of \(n\), such that \(a \le f(n) \le n\)? You might get a clue from concrete examples:

16 = 1 x 16 = 2 x 8 = 4 x 4;

24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6

or from considering the special case \(a=b\), where \(a = \sqrt{n}\).

It makes sense to try \(f(n) = \sqrt{n}\). Our examples work:

\(1, 2, 4 \le 4 = \sqrt{16}\)

\(1, 2, 3, 4 \le 4.9 \approx \sqrt{24}\)

Hence we are more confident of the **conjecture**:

*Let \(a \le b\) be any natural numbers. Let \(n = ab\). Then \(a \le \sqrt{n}\).*

You may wish to draw more examples of \(n\) squares arranged in a rectangle consisting of \(a\) rows and \(b\) columns, with \(a \le b\), and check that in each case \(a \le \sqrt{n}\). From such verifications, we are likely to believe the conjecture is true in general, i.e., we cannot find any natural numbers \(a \le b\) with product \(n\), such that \(a > \sqrt{n}\). However, as we have seen earlier, it is impossible to check every case, and a general argument is needed.

We now give a proof of the conjecture. From the first two statements, we know

(i) \(a \le b\).

(ii) \(n = ab\).

Now suppose the conclusion is false, i.e., \(a > \sqrt{n}\). Combining it with (i) gives \(b > \sqrt{n}\). Multiplying the two inequalities, we get \(ab > n\), which violates (ii). Assuming the conclusion is false has led to an absurdity. Therefore, the assumption is wrong, and the conclusion indeed follows from (i) and (ii). This concludes the proof. Now that the conjecture has been proved, it becomes a theorem.

The above is an example of **proof by contradiction**, which was used by the ancient Greeks to prove a variety of facts about numbers, such as there are infinitely many prime numbers, and the more surprising fact that \(\sqrt{2}\) is irrational, i.e., it cannot be expressed as the ratio of two natural numbers. The proofs of these facts are more intricate, but the underlying structure is exactly the same as here. In order to show that a conclusion follows from a premise, we assume that the conclusion is false.

Then we combine the premise with the assumption logically to derive a false statement. Something has gone wrong. Since the derivation is logically correct, we must give the assumption up. Hence the conclusion must follow from the premise.

As practice, you may want to write down a proof by contradiction to show that with the same premise as above, \(b \ge \sqrt{n}\). This fact also follows directly from the previous fact. From \(a \le \sqrt{n}\), we know \(1/a \ge 1/\sqrt{n}\).

Hence

[ b = n \times \frac{1}{a} \ge n \times \frac{1}{\sqrt{n}} = \sqrt{n} ]

We saw earlier that if \(a=b\), then both inequalities become equalities. Furthermore, if \(a<b\), then both inequalities are strict. In conclusion, we may make a more complete statement as follows: \

\noindent **Theorem (Product of Two Numbers).** {\it Let \(a \le b\) be any natural numbers, and let \(n = ab\).

Then

[ a \le \sqrt{n} \le b ]

Moreover, if \(a<b\), then \(a < \sqrt{n} < b\). Otherwise, \(a=b = \sqrt{n}\).

}

Let us remind ourselves that the conjecture and theorem are general statements, i.e., they are about all possible values of \(a\) and \(b\), as long as the values are natural numbers.

The phrase “any natural numbers” is a reminder.

We have come to the end of this entry.

Here are some other related problems.

(a) Formulate analogous conjectures about three natural numbers \(a, b, c\) and their product \(abc\), and then try to prove or disprove them.

(b) If you manage to prove a conjecture about three natural numbers, might you see a generalisation to the case where the product is formed from an unspecified number of natural numbers, say, \(k\), where \(k \ge 2\)?

(c) Does our Theorem hold if \(a\) and \(b\) are positive real numbers?