This exercise is very simple but it serves as a model for using \( (\neg q \Rightarrow \neg p ) \Leftrightarrow (p \Rightarrow q )\). That is to say, a model for proving something by proving the contrapositive of the statement. In other words to show \( p \Rightarrow q \) we will prove \( \neg q \Rightarrow \neg p\).
It is important to understand this type of argument because it is used often but is also almost always written words similar to “which is contrary to our assumption that…”. I find at times this language although correct can sounds suspicious. (When I don’t follow I sometimes think to myself “well who told you to assume that?”)
Because of its simplicity, this exercise is easy to follow and being in two parts, we can use this as model to fall back on. We’ll do the first part in somewhat pleonastic language, but the logical implications should be clearer. We’ll prove the second part (which has basically the same argument, only replacing addition with multiplication) with more usual language.
In this context in order to show that if \(x\) is irrational and \(r\) is rational (\(r\ne 0\)) then \(x+r\) is irrational, we will instead prove that if \(x+r\) is rational (i.e. not irrational) then \(x\) is rational (i.e. not irrational).
(a) If \(x\) is irrational and \(r \ne 0\) is rational then \(x+r\) is irrational.
Here is an overly formal way to write up our proof.
To prove the irrationality of \(x+r\) let’s assume the opposite (the negation.) So \(x+r\) is not irrational. This means that (we assume) \(x+r\) is rational.
Having assumed this, we now want to prove the negation of the original assumption (that \(x\) is not irrational.)
If \(x+r\) is rational then there exist integers \(p\) and \(q\) such that \(x+r=p/q\).
We also assume that \(r\) is rational, thus there also exists integers \(n\) and \(m\) such that \(r=n/m\).
We then have \(x+n/m=p/q\). Solving for \(x\) we see \(x = p/q-n/m = \frac{pm-nq}{qm}\). This is clearly a rational expression, thus \(x\) is rational, which is what we wanted to prove. We have shown that if \(x+r\) is not irrational (i.e. \(x+r\) is rational) then \(x\) is rational.
Therefore the contra-positive is true, if \(x\) is irrational then \(x+r\) is irrational.
(b) If \(x\) is irrational and \(r\ne 0\) is rational then \(xr\) is irrational.
We’ll now show the same for \(rx\), but will use the more usual vernacular:
Suppose \(x\) is irrational and \(r\) is rational (\(r \ne 0\)). If \(rx\) is rational then \(rx = n/m\) where \(n\) and \(m\) are integers (\(n, m \ne 0\)). Similarly since \(r\) is rational, \(r=p/q\). Thus \(\frac{p}{q}x=\frac{n}{m}\) and \(x=\frac{qn}{pm}\). This implies that \(x\) is rational, which contrary to our assumption that \(x\) is irrational.
Hence rational \(rx\) is impossible for irrational \(x\).