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You are watching: The sum of 2 rational numbers is rational

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.  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 $5+2\sqrt2$. For one, it comes directly from the closure of addition on $\altoalsimce.orgbb{Q}$, 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 $\altoalsimce.orgbb{Q}$ and $q_1, q_2 \neq 0$:$$a + b = \dfrac{p_1}{q_1} + \dfrac{p_2}{q_2} = \dfrac{p_1q_2 + p_2q_1}{q_1q_2} \in \altoalsimce.orgbb{Q}$$

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 $\altoalsimce.orgbb{Q}$ 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. 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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