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

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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

#### Answer:

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