Classification of Real Numbers: Difficult Problems with Solutions

Solution:
Collect the terms containing [tex]\sqrt{2}[/tex] on the left side:
[tex](a+2)\sqrt{2}=a-b-1[/tex]
If [tex]a+2\ne 0[/tex], then [tex]\sqrt{2}=\frac{a-b-1}{a+2}[/tex] would be a quotient of rational numbers, i.e. a rational number, which is false because [tex]\sqrt{2}[/tex] is irrational. Hence [tex]a+2=0[/tex], and then the right side is also zero:
[tex]a=-2,\qquad a-b-1=0\ \Rightarrow\ b=a-1=-3[/tex]
Therefore
[tex]a\times b=(-2)\times (-3)=6[/tex]

Solution:
Let [tex]x=0.999\ldots[/tex] (infinitely many nines). Multiplying by [tex]10[/tex] moves the decimal point one place to the right: [tex]10x=9.999\ldots[/tex]. After the decimal point both numbers have the same infinite tail of nines, so it cancels when we subtract:
[tex]10x-x=9.999\ldots-0.999\ldots=9\ \Rightarrow\ 9x=9\ \Rightarrow\ x=1[/tex]
This is the usual method for converting a repeating decimal into a fraction. The result can also be confirmed without operations on infinite decimals. For every natural number [tex]n[/tex], the number [tex]x[/tex] is at least the decimal with [tex]n[/tex] nines after the decimal point and at most [tex]1[/tex]:
[tex]1-\frac{1}{10^n}=0.\underbrace{99\ldots 9}_{n}\le x\le 1[/tex]
If [tex]x\ne 1[/tex], the number [tex]d=1-x[/tex] would be positive and [tex]d\le\frac{1}{10^n}[/tex] for every natural [tex]n[/tex]. This is impossible, because [tex]\frac{1}{10^n}\lt d[/tex] for [tex]n[/tex] large enough. Hence
[tex]0.999\ldots=1[/tex]

Solution:
Suppose, on the contrary, that [tex]\sqrt{3}[/tex] is rational. Since [tex]\sqrt{3}\gt 0[/tex], we can write [tex]\sqrt{3}=\frac{p}{q}[/tex], where [tex]p[/tex] and [tex]q[/tex] are natural numbers and the fraction is irreducible ([tex]p[/tex] and [tex]q[/tex] have no common divisor greater than [tex]1[/tex]). Squaring gives
[tex]p^2=3q^2[/tex]
Hence [tex]p^2[/tex] is divisible by [tex]3[/tex]. Then [tex]p[/tex] is also divisible by [tex]3[/tex]. Indeed, otherwise [tex]p=3k+1[/tex] or [tex]p=3k+2[/tex] for some integer [tex]k[/tex], and
[tex](3k+1)^2=3(3k^2+2k)+1,\qquad (3k+2)^2=3(3k^2+4k+1)+1[/tex]
In both cases [tex]p^2[/tex] leaves remainder [tex]1[/tex] when divided by [tex]3[/tex], a contradiction. So [tex]p=3m[/tex] for some natural number [tex]m[/tex], and
[tex]9m^2=3q^2\ \Rightarrow\ q^2=3m^2[/tex]
Hence [tex]q^2[/tex] is divisible by [tex]3[/tex] and, by the same argument, [tex]q[/tex] is divisible by [tex]3[/tex]. Thus [tex]3[/tex] divides both [tex]p[/tex] and [tex]q[/tex], which contradicts the irreducibility of the fraction [tex]\frac{p}{q}[/tex]. Therefore [tex]\sqrt{3}[/tex] is irrational.