Classification of Real Numbers: Very Difficult Problems with Solutions

Solution:
Let [tex]s=\sqrt{a}+\sqrt{b}+\sqrt{c}[/tex]; it is rational and [tex]s\gt 0[/tex]. We use that sums, differences, products and quotients (by a non-zero number) of rational numbers are rational.
Step [tex]1[/tex]: [tex]\sqrt{c}[/tex] is rational. We have [tex]\sqrt{a}+\sqrt{b}=s-\sqrt{c}[/tex]. Squaring both sides, we get
[tex]a+b+2\sqrt{ab}=s^2-2s\sqrt{c}+c\ \Rightarrow\ 2\sqrt{ab}=K-2s\sqrt{c}[/tex]
Here [tex]K=s^2+c-a-b[/tex] is a rational number. If [tex]K=0[/tex], then [tex]2\sqrt{ab}=-2s\sqrt{c}[/tex], which is impossible: the left side is positive and the right side is negative. Hence [tex]K\ne 0[/tex]. Squaring again gives [tex]4ab=K^2-4Ks\sqrt{c}+4s^2c[/tex], and since [tex]4Ks\ne 0[/tex],
[tex]\sqrt{c}=\frac{K^2+4s^2c-4ab}{4Ks}\in\mathbb{Q}[/tex]
Step [tex]2[/tex]: [tex]\sqrt{a}[/tex] and [tex]\sqrt{b}[/tex] are rational. The number [tex]u=\sqrt{a}+\sqrt{b}=s-\sqrt{c}[/tex] is rational and positive. Since [tex]\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b[/tex], the number [tex]\sqrt{a}-\sqrt{b}=\frac{a-b}{u}[/tex] is also rational. Therefore the following numbers are rational too:
[tex]\sqrt{a}=\frac{1}{2}\left(u+\frac{a-b}{u}\right),\qquad \sqrt{b}=\frac{1}{2}\left(u-\frac{a-b}{u}\right)[/tex]
This completes the proof. For example, it follows that [tex]\sqrt{2}+\sqrt{3}+\sqrt{5}[/tex] is irrational, because [tex]\sqrt{2}[/tex] is irrational.