*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 = x^{2}**.

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!**“.

I haven¦t checked in here for a while since I thought it was getting boring, but the last several posts are great quality so I guess I¦ll add you back to my daily bloglist. You deserve it my friend 🙂