Question on Irrationality of Irrational Numbers

Question

As we know that $2^{1/n}$ is an irrational number for all natural number greater than $1$.

But if we do $2^{n/n}$ then it will be a rational number or in simple words it will become a natural number.

Now as we know $\pi$ and $e$ is also irrational number then there also exist a number $k$ and $t$ which are natural numbers and the below relation will exist: $$ k^{1/q} = \pi \qquad \text{(for a unique value of $q$)}\\ t^{1/a} = e \qquad \text{(for a unique value of $a$)} $$

I know many people think about this problem but I don't know why we can't able to get the exact value of $k$, $q$, $t$ and $a$.

In this question I only want to know what are the limitation by which we don't getting these values.

Thanks for giving your time for this question.

Answer 1

Your mistake is in assuming that every irrational number is of the form $$a^{1/b}\tag{1}$$ for some natural numbers $a$ and $b$. This is not true. For instance, it could be easy for you to convince yourself that $$ \sqrt[3]{1+\sqrt{2}} $$ is

1. an irrational number, and
2. not of the form $a^{1/b}$ for any natural numbers $a$ and $b$.

---

Having convinced yourself of the above notion, you might feel inclined to loosen the restrictions on the type of irrational numbers you are interested in. After all, numbers of the form $a^{1/b}$ are precisely those that occur as roots of polynomials of the form $$ X^b - a = 0. $$ This is a polynomial with integer coefficients, and a very special polynomial too, clearly. It is intuitive to ask at this point whether $\pi$ or $e$ can be expressed as roots of some polynomial with integer coefficients. Perhaps $$ 5X^{10} - 3X^8 + 24 = 0, $$ or $$ 89X^{72} + 77 X^{55} - 55X^{33} - 123445 = 0 $$ are somehow the polynomials we seek? Clearly there are lots of irrational numbers that are being added here, and one can be hopeful that $\pi$ and $e$ land up in this collection.

As an aside, this collection is called the set of all algebraic numbers over the rationals.

---

Incredibly, the situation with $\pi$ and $e$ is even worse. There are no polynomials with integer coefficients that are satisfied by $\pi$ or by $e$! Numbers with this property are called transcendental. It is an elementary, but not an easy, thing to prove. If you have seen a bit of calculus then you could attempt to read Ivan Niven's monograph Irrational Numbers, which deals with these topics.