1. F bo 3 to [tex]\sqrt{9}[/tex] a 4 to [tex]\sqrt{16}[/tex] więc 4 jest bliższa liczbie [tex]\sqrt{13}[/tex]
2. P bo [tex]-1\frac{1}{4} =-\frac{5}{4}[/tex] a [tex]-0,8=-\frac{8}{10}=-\frac{4}{5}[/tex]
3. F bo [tex]\sqrt[3]{64} = 4[/tex] a [tex]\sqrt{4+21}=\sqrt{25}[/tex] czyli 5 lub -5 (+/- [tex]\frac{5}{1}[/tex] to liczba wymierna)
4. P bo [tex]0,875 = \frac{7}{8}[/tex]
