Absurdness of the Square Roots of few Fractions

Twice two makes four seems to me simply a piece of insolence. Twice two makes four is a pert coxcomb who stands with arms akimbo barring your path and spitting. I admit that twice two makes four is an excellent thing, but if we are to give everything its due, twice two makes five is sometimes a very charming thing too.

Fyodor Dostoevsky

Glad you came by. I wanted to let you know I appreciate your spending time here at the blog very much. I do appreciate your taking time out of your busy schedule to check out Math1089!

There are fractions of peculiar types which we shall consider here. When we take the square root of that fractions, an illusion takes place. Below is an example of such a mixed fraction.

There are a few things to observe in this fraction. First of all, the whole number 2 came out from the surd. Second, the whole number and the denominator (both 2 here) are same. Third, the denominator of the mixed fraction (here 3) is just 1 less than the square of the denominator (2 here). Consider one more example.

In this blogpost, we are looking for a rigorous solution for this question. First, look at the pattern. Our task is to find all such fractions satisfying the following relation

Squaring both sides, we have

Since x and y are natural numbers, cancellation yields y + 1 = x2.

In order to make x a natural number, possible values of x and y are given below:

  • y = 3, so that x = 2;
  • y = 8, so that x = 3;
  • y = 15, so that x = 4;
  • y = 24, so that x = 5;
  • y = 35, so that x = 6;
  • y = 48, so that x = 7;
  • y = 63, so that x = 8;

and so on and on.

Alternatively, from the above discussion we can convert the problem into a problem of single variable, say n. The fractions will satisfy the following relation involving n:

Of course, the value of n can be any natural number greater than 1. On putting various values of n we get various fractions. The validity of the result can be proved using the principle of mathematical induction.

Let P(n) be the mathematical statement

Then P(2) is true. Assuming P(n) is true, truth of P(n + 1) follows. In fact

and from LHS of P(n + 1) to RHS of P(n + 1) can be achieved as below.

Finally, we will consider few examples of the present problem.

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

%d bloggers like this: