i came throughout this inquiry in a book. I tried prove the condition as the following: suppose that a and also b room rational. Clearly the amount of $a$ and also $b$ is rational, which contradicts the condition, i m sorry is that $a+b$ is irrational. Therefore at the very least one the a or b is irrational.

You are watching: If a is a rational number and b is an irrational number, then the sum a + b is

I have actually a feeling that other is amiss there. Ns feel like using contrapositive proof here might be better but I"m no sure because I"m brand-new to the people of proofs.


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I desire to deal with your "I don"t see exactly how the contrapositive works here" comment.

Let $invernessgangshow.netbbI = invernessgangshow.netbbR setminus invernessgangshow.netbbQ$ (the set of irrational numbers).

You desire to display that

$$ a+b in invernessgangshow.netbbI implies a in invernessgangshow.netbbI vee b in invernessgangshow.netbbI$$

Before switching to the contrapositive, keep in mind that because that $a in invernessgangshow.netbbR$$$ lnot (a in invernessgangshow.netbbI) Leftrightarrow a in (invernessgangshow.netbbR setminus invernessgangshow.netbbI) Leftrightarrow a in invernessgangshow.netbbQ$$

Now, the contrapositive becomes

$$ lnot (a in invernessgangshow.netbbI vee b in invernessgangshow.netbbI) implies lnot (a+b in invernessgangshow.netbbI)$$which, in irradiate of the monitoring above, is$$ a in invernessgangshow.netbbQ land b in invernessgangshow.netbbQ implies a+b in invernessgangshow.netbbQ$$

which is a specifying property the $invernessgangshow.netbbQ$.

Remember likewise that $lnot (P vee Q) = (lnot P) land (lnot Q)$.


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answered Aug 17 "20 at 13:52
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AnalysisStudent0414AnalysisStudent0414
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The declare you"re trying to prove is $forall a,b, (a+b otin BbbQ implies a otin BbbQ ext or b otin BbbQ)$. This is just the symbolic translate in of the declare "for every $a,b$, if $a+b$ is irrational climate atleast among $a$ or $b$ is irrational".

Here, the declare $X$ is "$a+b otin BbbQ$", and also the explain $Y$ is "$a otin BbbQ ext or b otin BbbQ$". So, the contrapositive of "for every $a,b$ ($X implies Y$)" is "for every $a,b$ $( eg Y implies eg X)$", i beg your pardon in this situation is:

For every $a,b$ we have ($ain BbbQ$ and $bin BbbQ implies a+b in BbbQ$)

and this is what girlfriend argued.


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reply Aug 17 "20 at 13:55
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