Prove that for all #x,y in RR#, if #x# is rational and #y# is irrational, then #x+y# is irrational?

I know I need to use contradiction to prove this, but I'm not sure how to go about it. I know that for either to be rational, then #EE a,b in ZZ# such that #x=a/b# (for example) and #gcd(a,b)=1#. How can I go about proving this?

2 Answers
Feb 17, 2017

Answer:

Hint: Consider #(x+y)-x#

Explanation:

As is very often the case, we do not need to write this as a proof by contradiction. We can prove the contrapositive directly.

We can prove directly:

#x# is rational #rArr# (#x+y# is rational #rArr# #y# is rational)

(using #a,b in QQ rArr a-b in QQ# -- that is, #QQ# is closed under subtraction)

Therefore (by contraposition of the imbedded conditional)

#x# is rational #rArr# (#y# is not rational #rArr# #x+y# is not rational)

This is logically equivalent to

(#x# is rational #"&"# #y# is not rational) #rArr# #x+y# is not rational)

By contradiction

Suppose #p# and #not q# and #r# .

Prove a contradiction and conclude that if #p# and #not q#, then #not r#.

By contrapositive

Suppose #p# and #not r#. Prove that #q#.

Conclude that If #p# and #not q#, then #r#.

The two methods are very closely related and I don't know of anyone who accepts one and not the other. (Although many/most/all intuitionists refuse to accept either contradiction or contrapositive.)

Proof by contrapositive

Suppose that #x# is rational and #x+y# is rational.

Then the difference #(x+y) - x = y# is rational.

Hence is we know that #y# is irrational, then #x+y# must have been irrational. (Otherwise, #y# would have been rational after all.)

Feb 17, 2017

Proof by contradiction follows:

We have:

#x,y in RR " st " x in QQ# (the set of rational numbers), and
# " "y in {RR"\"QQ} # (the set of irrational numbers).

Let us assume that #x+y in QQ#, ie #x+y# is rational, then:

#EE m,n,p,q in ZZ # st #x=m/n# (since #x# is rational), and #x+y=p/q# (since the sum is rational).

Therefore, we can write;

# x + y =p/q #
# :. m/n+y=p/q #
# :. y=p/q - m/n #
# :. y=(np-mq)/(nq) #

And so #y# can be written as a fraction #=> y# is rational

But we initially asserted that #y# was irrational and hence we have a contradiction, and so the sum #x+y# cannot be rational and hence it must be irrational, QED.