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We call a number \(x\) rational if it can be represented as \(x=\frac{p}{q}\) for coprime integers \(p\) and \(q\). Otherwise we call the number irrational.
Non-zero numbers \(a\) and \(b\) satisfy the equality \(a^2b^2 (a^2b^2 + 4) = 2(a^6 + b^6)\). Prove that at least one of them is irrational.

Let the number \(\alpha\) be given by the decimal:

a) \(0.101001000100001000001 \dots\);

b) \(0.123456789101112131415 \dots\).

Will this number be rational?

Prove the irrationality of the following numbers:

a) \(\sqrt{3}{17}\)

b) \(\sqrt{2} + \sqrt{3}\)

c) \(\sqrt{2} + \sqrt{3} + \sqrt{5}\)

d) \(\sqrt{3}{3} - \sqrt{2}\)

e) \(\cos 10^{\circ}\)

f) \(\tan 10^{\circ}\)

g) \(\sin 1^{\circ}\)

h) \(\log_{2}3\)

Is it possible for

a) the sum of two rational numbers irrational?

b) the sum of two irrational numbers rational?

c) an irrational number with an irrational degree to be rational?

Prove that the number \(\sqrt {2} + \sqrt {3} + \sqrt {5} + \sqrt {7} + \sqrt {11} + \sqrt {13} + \sqrt {17}\) is irrational.

In the Republic of mathematicians, the number \(\alpha > 2\) was chosen and coins were issued with denominations of 1 pound, as well as in \(\alpha^k\) pounds for every natural \(k\). In this case \(\alpha\) was chosen so that the value of all the coins, except for the smallest, was irrational. Could it be that any amount of a natural number of pounds can be made with these coins, using coins of each denomination no more than 6 times?