Consider the following statements. Select all that are always true. The sum of a rational number and a rational number is rational. The sum of a ratio

a)Statement: The sum of a rational number and a rational number is rational. A rational number and a rational number are always rational when added together. For example, 
${\displaystyle \frac{1}{2}+\frac{2}{1}=\frac{1+4}{2\cdot 1}=\frac{5}{4}}$ 
Therefore, the statement is always true.
b)Statement: The sum of a rational number and an irrational number is irrational. The sum of a rational number and an irrational number is every time irrational.т For example, 
${\displaystyle \frac{1}{2}+(\frac{1}{2}+\sqrt{2})=1+\sqrt{2}}$ (irrational) 
Therefore, the statement is always true.
c)Statement: The sum of an irrational number and an irrational number is irrational. тThe sum of an irrational number and an irrational number is sometimes rational. For example, 
${\displaystyle (\frac{1}{2}+\sqrt{2})+\sqrt{2}=\frac{1}{2}+2\sqrt{2}}$ (irrational) 
and 
${\displaystyle (\frac{1}{2}+\sqrt{2})+(\frac{1}{2}-\sqrt{2})=\frac{1}{2}+\frac{1}{2}+\sqrt{2}-\sqrt{2}=1+0=1}$ (rational) 
Therefore, the statement is always false.
d)Statement: The product of a rational number and a rational number is rational. The product of two rational numbers is rational every time. For example, 
${\displaystyle \frac{1}{2}\cdot \frac{2}{1}=\frac{2}{2}}$ (rational) 
Therefore, the statement is always true.
e)Statement: The product of a rational number and an irrational number is irrational.The product of a rational number and an irrational number is every time irrational. For example, 
${\displaystyle \frac{1}{2}\cdot \sqrt{2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}}$ (irraational) 
Therefore, the statement is always true.
f)Statement: The product of an irrational number and an irrational number is irrational. The product of an irrational number and an irrational number is sometimes rational. For example, 
${\displaystyle \sqrt{2}\cdot \sqrt{3}=\sqrt{6}}$ (irrarional) 
and 
${\displaystyle \sqrt{3}\cdot \sqrt{27}=\sqrt{3\cdot 27}=\sqrt{81}=9}$ (rational) 
Therefore, the statement is false.