*Numbers are free creations of the human mind that serve as a medium for the easier and clearer understanding of the diversity of thought.*

**Richard Dedekind**

Welcome to the blog **Math1089 – Mathematics for All!**.

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From our early childhood we know that having two chocolates at disposal is a better proposition than having one! This indeed, tells about our intuitive acceptance of the order relation among positive integers (here 2 is more than 1). A similar ordering among all real numbers enables us to make statements about one real number being greater (or smaller) than the other.

For any two real numbers *x* and *y*, if *x* is less than *y* (written *x* < *y*) then *y* – *x* is positive. Geometrically, *x* < *y* if and only if *x* lies to the left of *y* on the real line. Likewise, *y* > *x* means *x* < *y*.

The case, either *x* = *y* and *x* < *y* can be described as *x* ≤ *y*. Similarly, *y* ≥ *x* means *x* ≤ *y*. We further have *x* > 0 if and only if *x* is positive and if *x* < 0, we say *x* is negative. However, for *x* ≥ 0 we say *x* to be non-negative.

For any two real numbers *x* and *y*, exactly one of the three relations *x* > *y*, *x* = *y*, *x* < *y* holds. It is the great *trichotomy law* of real numbers. It follows that, for any real number *x*, either *x* > 0 or *x* = 0 or *x* < 0.

If *x* ≠ 0, then *x*^{2} > 0 and *x*^{2} = 0 only when *x* = 0. It follows that *x*^{2} ≥ 0 for any real number *x*. Using this, we can prove that 1 > 0. In fact, 1 = 1^{2} and 1^{2} > 0. Hence 1 > 0. If *xy* > 0, then either both *x* and *y* are positive or, both *x* and *y* are negative. Finally, what if *x*^{2} < 0? In that case *x* is a complex number.

**Theorem 1. **If *a*^{2} + *b*^{2} = 0 for any two real numbers *a* and *b*, then *a* = 0 = *b* and conversely.

If *a* = 0 = *b*, then certainly *a*^{2} + *b*^{2} = 0 and we are done.

There are many ways to prove the other part, namely *a*^{2} + *b*^{2} = 0 means *a* = 0 = *b*. The above facts and results are important in proving this. The first proof is based on the law of trichotomy of real numbers.

By the law of trichotomy, either *a* > 0 or *a* = 0 or *a* < 0. In a similar fashion, we can have either *b* > 0 or *b* = 0 or *b* < 0. Collecting all the possibilities in the tabular form, we get the following table, from where we conclude that *a* = 0 = *b* is the only possibility for the present case.

Recall that, *x*^{2} = 0 means *x* = 0. When *a* ≠ 0 and *b* ≠ 0, we have *a*^{2} > 0 and *b*^{2} > 0. It follows that *a*^{2} + *b*^{2} > 0, an impossibility. Next suppose, *a* = 0 but *b* ≠ 0. This means, *a*^{2} = 0 and *b*^{2} > 0. It follows that *a*^{2} + *b*^{2} > 0, a contradiction. Finally, assume that *b* = 0 but *a* ≠ 0. Hence, *b*^{2} = 0 and *a*^{2} > 0, whence *a*^{2} + *b*^{2} > 0, a contradiction. The only possibility *a* = 0 = *b* is evident now.

Now *a*^{2} + *b*^{2} = 0 means *a*^{2} = − *b*^{2}. Since *a*, *b* are real numbers, we have *a*^{2}, *b*^{2} ≥ 0. If both of them are not equal to 0, let *a*^{2} = *m *> 0. Then, *b*^{2} = − *a*^{2} = − *m* < 0, contradicting the fact that *b*^{2} ≥ 0. Only possibility *a* = 0 = *b* follows.

We can write, from *a*^{2} + *b*^{2} = 0, that (*a* − *b*)^{2} + 2*ab* = 0 and (*a* + *b*)^{2} − 2*ab* = 0. Using the fact *x*^{2} ≥ 0 for all real number *x*, it follows that 2*ab* ≥ 0 and − 2*ab* ≥ 0. Hence *ab* ≥ 0 and *ab* ≤ 0. For both to hold simultaneously, only possibility is *a* = 0 = *b*.

Consider the points *Q*(*a*, *b*) and *P*(0, 0) in the plane. The distance between them is *PQ* = √(* a*^{2} + *b*^{2}). Now *a*^{2} + *b*^{2} =0 means *PQ* = 0. This is possible only when *Q* coincides with *P*, the origin. As a result (*a*, *b*) = (0, 0), whence *a* = 0 = *b* as required.

We now use complex numbers to establish the result. Recall that the product of two complex numbers is zero if and only if each one of them is zero. Now *a*^{2} + *b*^{2} = 0 means *a*^{2} − (−1)*b*^{2} = 0 or *a*^{2} − *i*^{2}*b*^{2} = 0. This in turn implies that (*a* + *ib*)(*a* − *ib*) = 0, the product of two complex numbers. Hence, either *a* + *ib* = 0 or *a* – *ib* =0. Equating real and imaginary parts on both sides, we get *a* = 0 = *b* in both cases. Therefore, only possibility for equality to hold is *a* = 0 = *b*.

**Few applications of the above result**

Of course, there is a similar generalization to the above theorem, which is given below.

**Theorem 2. **If *a*^{2} + *b*^{2} + *c*^{2} = 0 for any three real numbers *a*, *b* and *c*, then *a* = *b* = *c* = 0 and conversely.

Below, we will consider two examples which can be solved using the above theorem.

Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call **“****Math1089 – Mathematics for All!**“.