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 $lnot (P vee Q) = (lnot P) land (lnot Q)$.

re-superstructure

cite

follow

answered Aug 17 "20 at 13:52

AnalysisStudent0414AnalysisStudent0414

7,6671515 silver badges3535 bronze title

$endgroup$

include a comment |

0

$egingroup$

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.

re-superstructure

point out

monitor

reply Aug 17 "20 at 13:55

peek-a-boopeek-a-boo

28.5k22 yellow badges1717 silver- badges4646 bronze badges

$endgroup$

2

include a comment |

## your Answer

Thanks for contributing an answer to invernessgangshow.net Stack Exchange!

Please be certain to*answer the question*. Administer details and share your research!

But *avoid* …

Use invernessgangshow.netJax to style equations. invernessgangshow.netJax reference.

To find out more, view our tips on writing good answers.

See more: Hatchback/Liftgate Hinge Problems Of Ford Explorer Rear Window Hinge Recall

Draft saved

Draft discarded

### Sign up or log in

sign up making use of Google

authorize up making use of Facebook

sign up utilizing Email and Password

send

### Post together a guest

surname

email Required, however never shown

### Post together a guest

surname

Required, yet never shown

short article Your price Discard

By clicking “Post your Answer”, you agree come our terms of service, privacy policy and cookie plan

## Not the answer you're looking for? Browse various other questions tagged solution-verification or ask your very own question.

The Overflow Blog

Featured on Meta

Linked

2

If $a$ and also $b$ are both irrational, is $a+b$ also irrational?

associated

2

Proof that irrational number are thick

2

proof by contrapositive: Prove for all $x,yininvernessgangshow.netbbR,$ if $x$ is rational and also $y$ is irrational climate $x+y$ is irrational.

2

Prove by induction the $3^4n+2+1$ is divisible by $5$ when $n ge 0.$

1

Proving the if $invernessgangshow.netrmker, RTR^-1= invernessgangshow.netrmker, RSR^-1$ then $invernessgangshow.netrmker T = invernessgangshow.netrmkerS$.

1

Proving the it's difficult to prove irrationality of all genuine numbers.

4

just how to prove the if c is an odd integer that divides the sum and the distinction of 2 integers a and b, climate c divides both a and also b?

warm Network inquiries more hot inquiries

question feed

subscribe to RSS

concern feed To subscribe to this RSS feed, copy and also paste this URL into your RSS reader.

invernessgangshow.net

company

stack Exchange Network

site architecture / logo design © 2021 stack Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.10.28.40592

invernessgangshow.netematics stack Exchange works finest with JavaScript enabled

your privacy

By clicking “Accept all cookies”, you agree stack Exchange deserve to store cookies on your an equipment and disclose information in accordance v our Cookie Policy.