Who all constitute algebraic numbers?

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Algebraic numbers are numbers which are roots of polynomial with rational coefficients. What all constitutes these numbers. Transcendental numbers are those which can not be roots of above. And wiki says ex as $\pi$ and e

Can you pls elaborate on this what all are included under algebraic number?

  1. square root of prime? (My thinking $x^2 - p = 0$ , p is prime) so all square root of prime are algebraic

  2. all complex numbers - wiki says yes (My thinking - $x^2+1 = 0$ )

**3. All irrationals? -- i could think of some irrationals like $x^2-2 = 0$. How to conclude all irrationals as algebraic?**All irrationals are algebraic? iirational means non-recurring non-terminating decimal number.

  1. All rationals? - I think yes $ xa+b = 0$ and $a$ and $ b$ are co-prime right?

  2. All primes? --> (my thinking - x-p = 0, p is prime)

  3. Any other?

Some may think that it is stupid Qn, may be yes. But before downvoting or recommending to delete it, atleast enlighten me! I am a beginner and learning on my own. So i have posted here to clarify my ignorance. pardon me for being so stupid.


can u please elaborate on what kind of irrationals make them not algebraic? (i know only pi and e) . I got this question when i was thinking on $\sqrt p, p$ a prime number set is subset of irrationals(which are uncountable) yet countable. But on other side algebraic numbers are countable.

That made me to think what subset of irrationals are not countable. So went to wiki for seeing algebraic numbers and rest all here.

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Actually, algebraic numbers are numbers which are roots a a polynomial in $\mathbf Z[X]$. Among these, algebraic integers are numbers which are root of a monic polynomial in $\mathbf Z[X]$.

Therefore the answer to 1, 4 and 5 is yes (albeit, in the case of rationals, it has nothing to do with $a$ and $b$ being coprime).

The answer to 2 is No: some complex numbers are algebraic, others are not. Anyway, remember a real number is also a complex number. For instance $\pi i$ is transcendental.

The answer to 3 is also No. Actually the irrationals are by definition the numbers which are not rational. So either they're algebraic or they're transcendental.