The brief answer come your concern is the is no necessarily true. Because that instance, $\\sqrt2, \\sqrt2-1,1-\\sqrt2$ are all irrational yet $$\\sqrt2 + (1-\\sqrt2) = 1 \\in \\nadechworld.combbQ$$ and $$\\sqrt2 - (\\sqrt2-1) = 1 \\in \\nadechworld.combbQ$$

However, the is worth noting that if $x$ and $y$ are irrational, then either $x+y$ or $x-y$ is irrational i.e. $x+y$ and $x-y$ can not be both rationals. The proof for this is given below.

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Proof

If both $x+y$ and $x-y$ are rational, climate we have that $x+y = \\dfracp_1q_1$ and also $x-y = \\dfracp_2q_2$, where $p_1,p_2 \\in \\nadechworld.combbZ$ and also $q_1,q_2 \\in \\nadechworld.combbZ \\backslash \\0\\$.

Hence, $x = \\dfrac\\dfracp_1q_1 + \\dfracp_2q_22 = \\dfracp_1q_2 + p_2q_12q_1q_2$ and also $y = \\dfrac\\dfracp_1q_1 - \\dfracp_2q_22 = \\dfracp_1q_2 - p_2q_12q_1q_2$.

Now $p_1q_2 + p_2q_1, p_1q_2 - p_2q_1 \\in \\nadechworld.combbZ$, whereas $2q_1q_2 \\in \\nadechworld.combbZ \\backslash \\0\\$.

This contradicts the fact that $x$ and also $y$ room irrationals. Hence, if $x$ and $y$ space irrational climate either $x+y$ is irrational or $x-y$ is irrational.

Below space some statements precious knowing.

1 sum of 2 rationals is always a rational.

Proof: permit the two rationals be $\\dfracp_1q_1$ and $\\dfracp_2q_2$, whereby $p_1,p_2 \\in \\nadechworld.combbZ$ and $q_1,q_2 \\in \\nadechworld.combbZ \\backslash0$. Then $$\\dfracp_1q_1 + \\dfracp_2q_2 = \\dfracp_1q_2 + p_2q_1q_1q_2$$where $p_1q_2 + p_2q_1 \\in \\nadechworld.combbZ$ and also $q_1 q_2 \\in \\nadechworld.combbZ \\backslash \\0\\$. Hence, the amount is again a rational.

2 sum of a rational and an irrational is always irrational.

Proof: allow the reasonable number be of the type $\\dfracpq$, wherein $p \\in \\nadechworld.combbZ$ and $q \\in \\nadechworld.combbZ \\backslash \\0\\$ while the irrational number it is in $r$. If $r + \\dfracpq$ is a rational, climate we have that $r + \\dfracpq = \\dfracab$ for part $a \\in \\nadechworld.combbZ$ and also $b \\in \\nadechworld.combbZ \\backslash \\0\\$. This means that $r = \\dfracab - \\dfracpq = \\dfracaq-bpbq$ wherein $aq-bp \\in \\nadechworld.combbZ$ and $bq \\in \\nadechworld.combbZ \\backslash \\0\\$. This contradicts the reality that $r$ is irrational. Hence, our assumption that $r + \\dfracpq$ is a reasonable is false. Hence, $r + \\dfracpq$ is a irrational.

3 amount of 2 irrationals have the right to be rational or irrational.

Example for sum of two irrationals gift irrational

$\\sqrt2$ is irrational. $\\sqrt2 + \\sqrt2 = 2 \\sqrt2$ i m sorry is again irrational.

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Example for sum of two irrationals being rational

$\\sqrt2$ and $1-\\sqrt2$ room irrational. (Note that $1-\\sqrt2$ is irrational indigenous the 2nd statement.) But, $\\sqrt2 + (1-\\sqrt2) = 1$ i beg your pardon is rational.