1

I really don't know how to prove it. I can prove that $\sqrt[3]{4}$ is irrational and prove that $\sqrt{3}$ is irrational. but as we know , sum of 2 irrational can be irrational or rational ($\sqrt{2} + -\sqrt{2} =$ rational). so I tried to prove that the sum of 2 different positive irrational numbers is always irrational but also failed.

anyone know?

Noa Even
  • 2,801
Roach87
  • 696
  • 3
    The sum of two different positive irrational numbers can be rational, e.g. $\sqrt{2}+(2-\sqrt{2})=2$. – J.G. Jul 23 '20 at 11:44
  • Yes I know , thats why I failed to prove that the sum of 2 irrational is irrational. – Roach87 Jul 23 '20 at 11:45
  • 1
    Hint: Suppose $\sqrt[3] 4=\alpha-\sqrt 3$ for some $\alpha \in \mathbb Q$. Now cube both sides. – lulu Jul 23 '20 at 11:45

2 Answers2

7

Let $x=\sqrt{3}+\sqrt[3]{4}$ so $0=(x-\sqrt{3})^3-4=x^3-3x^2\sqrt{3}+9x-3\sqrt{3}-4$. If $x\in\Bbb Q$, $\sqrt{3}=\frac{x^3+9x-4}{3(x^2+1)}\in\Bbb Q$, a contradiction.

J.G.
  • 115,835
0

The following should allow you to generalize to combinations of a quadratic and a cubic irrational in general:

Every number of the form $\alpha=a+b\sqrt 3$ with $a,b\in\Bbb Q$ is a root of a quadratic polynomial because $(a+b\sqrt3)^2=a^2+3b^2+2ab\sqrt 3$ and so $\alpha$ is a solution of of $$X^2-2aX+a^2-3b^2=0. $$ If $\sqrt 3-\sqrt[3]4$ is rational, it follows that the above applies to $\alpha:=\sqrt[3]4$. But we also know that it is a solution of the $X^3-4=0$. Hence it is also a solution of $$ 0=(X^3-4)-(X+2a)\cdot(X^2-2aX+a^2-3b^2)=(\ldots)\cdot X+(\ldots)$$ and therefore rational - contradiction. (Verify that a least one of the "$(\ldots)$" is non-zero!)

Hagen von Eitzen
  • 374,180