From Wikipedia

A

transcendental numberis a real or complex number that is not algebraicA

transcendental functionis an analytic function that does not satisfy a polynomial equationHowever these definitions are arguably rather cryptic to those who are not familiar with the literature of higher mathematics.

So in layman’s terms, what exactly does it mean to be transcendental? How would a

transcendental numberbe different from an ordinary number, say 5. And respectively, how would atranscendental functionbe different from an ordinary function, say f(x) = x^2

**Answer**

We will play a game. Suppose you have some number x. You start with x and then you can add, subtract, multiply, or divide by any integer, except zero. You can also multiply by x. You can do these things as many times as you want. If the total becomes zero, you win.

For example, suppose x is \frac23. Multiply by 3, then subtract 2. The result is zero. You win!

Suppose x is \sqrt[3] 7. Multiply by x, then by x again, then subtract 7. You win!

Suppose x is \sqrt2 +\sqrt3. Here it’s not easy to see how to win. But it turns out that if you multiply by x, subtract 10, multiply by x twice, and add 1, then you win. (This is not supposed to be obvious; you can try it with your calculator.)

But if you start with x=\pi, you *cannot* win. There is no way to get from \pi to 0 if you add, subtract, multiply, or divide by integers, or multiply by \pi, no matter how many steps you take. (This is also not supposed to be obvious. It is a very tricky thing!)

Numbers like \sqrt 2+ \sqrt 3 from which you can win are called

algebraic. Numbers like \pi with which you can’t win are calledtranscendental.

Why is this interesting? Each algebraic number is related arithmetically to the integers, and the winning moves in the game show you how so. The path to zero might be long and complicated, but each step is simple and there is a path. But transcendental numbers are fundamentally different: they are *not* arithmetically related to the integers via simple steps.

**Attribution***Source : Link , Question Author : AlanSTACK , Answer Author :
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