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.
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$\begingroup$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 $5+2\sqrt2$.
$\endgroup$ 2 $\begingroup$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.)
$\frac pq$+$\frac xz$ $(q,z \neq 0)$(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.
$\endgroup$ 0 $\begingroup$The sum of two irrational numbers is not necessarily irrational. For example, $\sqrt{2}$ and $-\sqrt{2}$ are two irrational numbers, but their sum is zero ($0$), which in turn is rational.
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