The sum of squares is zero then each of them is separately Zero

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

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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 x² > 0 and x² = 0 only when x = 0. It follows that x² ≥ 0 for any real number x. Using this, we can prove that 1 > 0. In fact, 1 = 1² and 1² > 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 x² < 0? In that case x is a complex number.

Theorem 1. If a² + b² = 0 for any two real numbers a and b, then a = 0 = b and conversely.

If a = 0 = b, then certainly a² + b² = 0 and we are done.

There are many ways to prove the other part, namely a² + b² = 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, x² = 0 means x = 0. When a ≠ 0 and b ≠ 0, we have a² > 0 and b² > 0. It follows that a² + b² > 0, an impossibility. Next suppose, a = 0 but b ≠ 0. This means, a² = 0 and b² > 0. It follows that a² + b² > 0, a contradiction. Finally, assume that b = 0 but a ≠ 0.  Hence, b² = 0 and a² > 0, whence a² + b² > 0, a contradiction. The only possibility a = 0 = b is evident now.

Now a² + b² = 0 means a² = − b². Since a, b are real numbers, we have a², b² ≥ 0. If both of them are not equal to 0, let a² = m > 0. Then, b² = − a² = − m < 0, contradicting the fact that b² ≥ 0. Only possibility a = 0 = b follows.

We can write, from a² + b² = 0, that (a − b)² + 2ab = 0 and (a + b)² − 2ab = 0. Using the fact x² ≥ 0 for all real number x, it follows that 2ab ≥ 0 and − 2ab ≥ 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 = √( a² + b²). Now a² + b² =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 a² + b² = 0 means a² − (−1)b² = 0 or a² − i²b² = 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 a² + b² + c² = 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.

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