Question

choose whether each statement is \always,\ \sometimes,\ or
ever\ true.

statement	true?
the sum of an irrational number and an irrational number is irrational.	always sometimes never
the product of a rational number and a rational number is rational.	always sometimes never
the product of an irrational number and an irrational number is irrational.	always sometimes never

Explanation:

Step1: Analyze the sum of rational and irrational

Let \( r \) be rational (\( r = \frac{a}{b}, a,b\in\mathbb{Z}, b
eq0 \)) and \( i \) be irrational. Assume \( r + i = s \) is rational. Then \( i = s - r \), but the difference of two rationals is rational, contradicting \( i \) irrational. So sum is always irrational.

Step2: Analyze the sum of two irrationals

Take \( \sqrt{2} \) (irrational) and \( -\sqrt{2} \) (irrational): their sum is \( 0 \) (rational). Take \( \sqrt{2} \) and \( \sqrt{3} \): sum is irrational. So sum is sometimes irrational.

Step3: Analyze the product of two rationals

Let \( r_1=\frac{a}{b}, r_2=\frac{c}{d} \) (\( a,b,c,d\in\mathbb{Z}, b,d
eq0 \)). Product is \( \frac{ac}{bd} \), which is rational (since \( ac, bd\in\mathbb{Z}, bd
eq0 \)). So product is always rational.

Step4: Analyze the product of two irrationals

Take \( \sqrt{2} \) and \( \sqrt{2} \): product is \( 2 \) (rational). Take \( \sqrt{2} \) and \( \sqrt{3} \): product is \( \sqrt{6} \) (irrational). So product is sometimes irrational.

Answer:

Statement	True?
The sum of an irrational number and an irrational number is irrational.	Sometimes
The product of a rational number and a rational number is rational.	Always
The product of an irrational number and an irrational number is irrational.	Sometimes