How do I prove that the following proposition is true?
The sum of any rational number and any irrational number is irrational.
I am currently a beginner at discrete math and I am still getting used to the format of writing proofs.
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Xander Henderson
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Marieeee
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1Does this answer your question? Is the sum and difference of two irrationals always irrational? – Mrcrg Jul 10 '20 at 16:10
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1The difference of two rational numbers is always rational. See $\frac{z_1}{z_2}-\frac{z_3}{z_4}=\frac{z_1z_4-z_3z_2}{z_2z_4}$, where all the $z_i$s are integers. If the sum of an irrational and rational were rational, then there exists an irrational that can be expressed as the difference of two rationals- a contradiction. – Michael Ejercito Feb 21 '24 at 04:39
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Say we have $r\in\mathbb{R}\setminus\mathbb{Q}$ and $s\in\mathbb{Q}$. Now assume by contradiction that $r+s\in\mathbb{Q}$. Therefore, there exists such $q\in\mathbb{Q}$ that satisfies $r+s=q$.
Now let's transfer s. $\mathbb{Q}$ is a field, so $q,s\in\mathbb{Q}$ implies $r=q-s\in\mathbb{Q}$, and that's a contradiction to the initial assumption that $r\notin\mathbb{Q}$
GBA
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I think the somewhat more usual notation for "the set of irrationals" is $\Bbb R \setminus \Bbb Q$, produced with "
\setminus". – Izaak van Dongen Jul 10 '20 at 17:00
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Let $r$ be rational and let $i$ be irrational.
Now assume that $s:=r+i$ is rational.
Subtractiong $r$ on both sides we find $s-r=i$.
But $s$ and $r$ are both rational and it is well known that in that case $s-r$ is a rational number.
But also $s-r=i$ hence is irrational.
This cannot go together so actually we deduced a contradiction.
Then we are allowed to conclude that our assumption is false which means exactly that $s$ is irrational.
drhab
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