Is the sum of two rationals or two irrationals irrational?

Question

1. I know this statement is false (if I am correct) but how to prove it's false?

"The sum of two rational numbers is irrational."

2. I know this statement is true (if I am correct) but how to prove it's true?

"The sum of two irrational numbers is irrational"

I used the example $\sqrt{2}+ \sqrt{3} = 3.14$

But i may need to use proof by contradiction or contaposition.

Answer 1

If two numbers are rational we can express their sum as $$\frac{a}{b} + \frac{c}{d}$$ which is equal to $$\frac{ad + bc}{bd}.$$ Hence, rational.

The sum of two irrational numbers may be irrational. Consider $2+\sqrt2$ and $3+\sqrt2$. Both are irrational, and so is their sum $5+2\sqrt2$.

Answer 2

For one, it comes directly from the closure of addition on $\mathbb{Q}$, but I don't think that's the answer they would expect.

Let $a = \dfrac{p_1}{q_1}$ and $b = \dfrac{p_2}{q_2}$ be rationals in $\mathbb{Q}$ and $q_1, q_2 \neq 0$: $$a + b = \dfrac{p_1}{q_1} + \dfrac{p_2}{q_2} = \dfrac{p_1q_2 + p_2q_1}{q_1q_2} \in \mathbb{Q}$$

For the second one, how about $\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} = \sqrt{2}$. A single example is sufficient to prove the claim.

For bonus points, can you prove that $\dfrac{\sqrt{2}}{2}$ is irrational?
(Hint: Contradiction. Suppose it's rational, and use the closure of addition on $\mathbb{Q}$ that was proven.)

Answer 3

$\frac pq$+$\frac xz$ $(q,z \neq 0)$(by formula of rational numbers).

=$\frac{pz+qz}{qz}$ ,which is again in the form $\frac ab$ so it is bound to be rational and also $qz$ is not equal to $0$.

Sum of irrational may be irrational is true but it is always rational if the sum consists of the irrational number and its negative and then the sum will yield $0$. Sum of two irrational numbers that you expressed as a decimal is not true and only an approximation.