Which One is Bigger – A or B?

Complete knowledge of the nature of an analytic function must also include insight into its behavior for imaginary values of the arguments. Often the latter is indispensable even for a proper appreciation of the behavior of the function for real arguments. It is therefore essential that the original determination of the function concept be broadened to a domain of magnitudes which includes both the real and the imaginary quantities, on an equal footing, under the single designation complex numbers.

Carl Friedrich Gauss

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Bigger or smaller, as we all know, that entire mathematics is full of such questions. One gets an intuitive idea of bigger and smaller, probably at the age of around 5 years or so. 1 is smaller than 2 or 5 is bigger than 4 are appropriate for this stage. As we grow up, difficulty level increases and eventually we are asked to compare the numbers like 1001 and 10001! As the passage of time, the turn comes to compare 2³ and 3². This is the case when the student is studying in class 7.

While studying in senior secondary level, the student has to face the same kind of questions with different inputs. Of course, they study complex numbers in these classes. Yes, you are correct. In this article, we will consider an important example from complex numbers.

Suppose A = 9 + 8i and B = 2 + 3i, where i² = −1 are two complex numbers. Definitely, the real parts are 9 and 2; imaginary parts are 8 and 3. Of course, no wonder that 9 > 2 and 8 > 3. Anyway, can we conclude A > B?

As long as we are in the real domain, of course, exactly one of the three relations is true: either x > y or x = y or x < y. We also know that, if x is real then x² ≥ 0 and x² = 0 if and only if x = 0. Great, that’s enough to explain the given example.

One major difference about the complex numbers in comparison to the purely real numbers is that the usual ordering (means the sense of greater than or less than). This is a powerful tool in reals but is meaningless in the system of complex numbers. However, the present statement makes no sense among complex numbers, unless both the numbers are purely real.

Before we conclude that A is bigger than B or B is bigger than A, the key point is to know whether i > 0 or i < 0. Consider the complex numbers i and 0. Clearly, i ≠ 0. For if i = 0, then i² = 0 which contradicts the fact that i² = −1. Since i ≠ 0, either i > 0 or i < 0. Now see the next.

If possible, let i > 0. Then i × i > 0 × 0 or i² > 0. This means that

 − 1 > 0,                                [A]

or, – 1 + 1 > 0 + 1,             [adding 1 to both sides]

or, 0 > 1.                              [X]

Of course, this is not possible in the usual ordering. May be to explore some new order among the complex numbers, we need to proceed further. Let us further investigate this possibility. We have

− 1 > 0,                                 [A]

or, (– 1) (– 1) > (0)(0)

or, 1 > 0.                              [Y]

Clearly, both [X] and [Y] cannot hold simultaneously. Hence i > 0 is not possible.

Next, let us assume that i < 0. In this case too, i × i > 0 or i² > 0. This means that −1 > 0, whence 0 > 1 as shown above. Considering this − 1 > 0, and following the similar lines of argument as shown above, we can show that 1 > 0. Of course, 0 > 1 and 1 > 0 cannot hold simultaneously. Hence i < 0 is also impossible.

It follows that, neither i = 0 nor i < 0 nor i > 0. So, we sum up with no meaning assigned to the statements like a + ib > c + id (or a + ib < c + id), where a + ib and c + id are any imaginary numbers.

In the present situation, given that A = 9 + 8i and B = 2 + 3i. Of course, we cannot conclude that A > B. We conclude the article after considering the following problems.

1. Is it true that 8 + 9i > 3 – 2i?

2. Is it true that i < 2i?

3. For any two real numbers a and b, the statement ai > bi is

(a) true only if a > b         (b) true only if |a| > b    (c) true only if |a| > |b|                 (d) never true.

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