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?
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
2 Answers
Hint: Consider
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:
(using
Therefore (by contraposition of the imbedded conditional)
This is logically equivalent to
(
By contradiction
Suppose
Prove a contradiction and conclude that if
By contrapositive
Suppose
Conclude that If
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
Then the difference
Hence is we know that
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
#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
But we initially asserted that