1

Prove the following:

The product of a nonzero rational number and an irrational number is also irrational.

I assumed the following:

Let $r = c/d$ be rational, where $c$ and $d$ are integers and $r$ is nonzero, so $c$ and $d$ are nonzero as well. Let $i$ be irrational.

Then I tried proving by contradiction that:

Suppose that $ri = a/b$. Then $(ci)/d = a/b$. I assume that $ci$ is also irrational because multiplying an irrational number by an integer results in an irrational number (but I don't know why this is, to be honest).

Is that correct?

Community
  • 1
daggerissues
  • 21
  • Keep going. Divide both sides by $c$ and mulitply both sides by $d$. to get $i = \frac{ad}{bc}$.... If $i$ is irrational and $c\ne 0$ is and integer then $ci$ is irrational because if $ci = r$ we're ratioanal then $i = \frac rc$ would be rational divided by an integer, which is rational. Why. Because if $r= \frac ab$ then $\frac rc = \frac {a}{bc}$ and $a, bc$ are integers. – fleablood Apr 11 '20 at 21:39

3 Answers3

0

You have to go a bit further and show that, with your assumptions, $i$ would be a rational number $\frac{ad}{cb}$, thus showing the contradiction. ($ci$ is indeed irrational, but this fact is not needed for the proof.)

Parcly Taxel
  • 103,344
  • Thank you for the clarification. I actually did lay out a line like that but wasnt sure if it was worth adding. I tried verifying with algebra but it led me in circles. What you say makes sense. Appreciate your time! – daggerissues Apr 11 '20 at 18:36
  • @daggerissues Please upvote and accept my answer since you found it useful. – Parcly Taxel Apr 11 '20 at 18:36
0

We only need to know that the rationals form a field: Suppose that $ri$ is rational (i.e. $ri \in \Bbb Q$)
then $i= \frac{ri}{r} \in \Bbb Q$ because we can divide a rational by a non-zero rational and the result is rational. But $i$ is given not to be rational. This contradiction shows $ri$ is not rational, so irrational.

Henno Brandsma
  • 242,131
0

If $ri = \frac ab$ where $a,b$ are integers and $r = \frac cd\ne 0$ then $c \ne 0$ and

$ri = \frac cdi = \frac ab$ and

$i = \frac {ad}{cd}$ which is rational.

====

Basic props.

Prop 1: If $r$ is rational and $n$ is an integer then $nr$ and $r+n$ and $r-n$ and, if $n \ne 0$ then $\frac rn$ are rational.

Pf: If $r$ is rational then there are integers, $a,b$ so that $r= \frac a b$. So $nr = \frac {an}b$ and $r + n = \frac {a + bn}n$ and $r-n = \frac {a-bn}b$ and $\frac rn = \frac a{bn}$ and as $an, a+bn, a-bn,$ and $bn$ are all integers those are all rational.

Prop 2: If $r$ is rational and $q$ is rational then $qr$ and $q+r$ and $q-r$ and, if $q\ne 0$ the $\frac rq$ are rational..

Pf: $r$ is rational then $r =\frac ab$ form some integers $a,b$ and $q=\frac cd$ for some integers $c,d$. $qr = \frac {ac}{bd}, q+r = \frac{ad + bc}{bd}, q-r = \frac{ad-bc}{bd}$ and $\frac rq = \frac {ad}{bc}$ and as $ac,bd,ad+bc, ad-bc, ad,$ and $bc$ are all integers those are all rational.

Corollary: If $r$ is rational and $q$ is rational and if $xr = q$ or $\frac rx = q$ or $\frac rx = q$ or $r+x=q$ or $r-x = q$. If any of those hold then $x$ is rational.

Pf: If any of those hold then $x = \frac qr$ or $x=\frac rq$ or $x=q-r$ or $x = r-q$. And by 1, and 2) above those are all rationals.

Prop: If $i$ is irrational and $p$ is rational then $pi, \frac ip$ (assuming $p\ne 0$), $\frac pi, p+i, p-i, i-p$ etc are all irrational. Pf: If any of the results were rational the the corolary above would prov to us that $i$ is rational. Which it isn't.

THE ONE EXCEPTION.

If $i$ is irrational and $j$ is irrational, then we have utterly NO way of knowing whether $ij, \frac ij, \frac ji, i+j, i-j, j-i$ are rational or irrational or what.

fleablood
  • 124,253