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.

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

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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