# Proving that $\sqrt 7 -\sqrt 2$ is irrational.

I understand proving that $$\sqrt{7}-\sqrt {2}$$ is irrational, but how does the answer change if its cube root of $$7$$ instead of square root?

the way I solve $$\sqrt{7}-\sqrt {2}$$ is by assuming its rational, and multiplying both sides by $$\sqrt{7}+\sqrt {2}$$, and deriving a contradiction

Thank you!

• Sketch: say it is rational. So $\sqrt 2 = r +\sqrt 7$. Square and derive a contradiction. Or write $\sqrt 7 = s+\sqrt 2$, cube and derive a contradiction (that way is faster).
– lulu
Apr 16, 2021 at 19:27
• @user904299 Could you expand a bit on the method you know to prove that $\sqrt{7} - \sqrt{2}$ is irrational which would probably allow us to tailor an answer better? Apr 16, 2021 at 19:31
• @lulu is s going to be in the form m/n? (where both are integers, and we go from there) Apr 16, 2021 at 19:31
• If you read what the original poster wrote, you can understand that actually he wanted to write $\sqrt7-\sqrt2$, indeed he wrote “how does the answer change if its square root of $7$ instead of cube root?” Apr 16, 2021 at 19:32
• Very similar problem: Prove that $\sqrt{5} + \sqrt{2}$ is irrational. Apr 16, 2021 at 19:37

Let’s go for an elementary proof (no rational root theorem and no field extension)

Assume $$r=\sqrt{7}-\sqrt{2}$$ is rational. This means $$\sqrt{7}=r+\sqrt{2}$$. Cubing one gets

$$7=r^3+3r^2\sqrt{2}+6r+2\sqrt{2}$$

$$\sqrt{2}={7-r^3-6r\over 3r^2+2}$$

$$r$$ being rational this means $$\sqrt{2}$$ is rational. A contradiction

$$\mathbb{Q}(\sqrt{7})$$ is a degree 3 extension of $$\mathbb{Q}$$.

Suppose that $$\sqrt{7} - \sqrt{2}$$ is rational (call it $$a$$). Then, $$\sqrt{7} = \sqrt{2} + a$$. Therefore, $$\mathbb{Q}(\sqrt{7}) = \mathbb{Q}(\sqrt{2})$$.

However, $$\mathbb{Q}(\sqrt{2})$$ is a degree 2 extension of $$\mathbb{Q}$$. This is a contradiction, as the same extension cannot be both degree 2 and degree 3.

Let $$\alpha=\sqrt 7 -\sqrt 2$$.

Then $$\alpha$$ is a root of $$x^6 - 6 x^4 - 14 x^3 + 12 x^2 - 84 x + 41=0$$. Since this is a monic polynomial with integer coefficients (the actual polynomial is not important), the rational root theorem tells you that $$\alpha$$ is either irrational or an integer.

Now

$$\quad 1.9 <\sqrt 7 < 2.0$$

$$\quad 1.4 < \sqrt 2 < 1.5$$

gives

$$\quad 0.4 < \alpha < 0.6$$

which proves that $$\alpha$$ is not an integer and so must be irrational.

We can avoid these fine estimates: if $$\alpha$$ is an integer then it must divide $$41$$ (now the polynomial is important), but clearly $$0 < \alpha < \sqrt 7 < 2$$ and $$\alpha \ne 1$$.