Question

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?

Answer 1

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.)

Answer 2

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.