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I am having trouble working about a proof for the claim: If a and b are any rational numbers, with b not equal to 0, and r is any irrational number, then a+br is irrational.

This is all I really have so far...

Proof: 

Let rational numbers a and b as well as irrational number r be given.
Suppose b is not equal to 0
Since b is not equal to 0, there exists an integer k such that b=k+1

that really where I am stuck and I'm not even sure that the last line there even makes sense. If anyone could help guide me along that would be wonderful! Thanks.

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    So the last line does not make sense because $k$ is an integer and $1$ an integer, so $k+1$ is an integer and thus $b$ is an integer, which is not always the case if we only know that $b$ is rational. I would suggest arguing this by contradiction. Assume that there exists some irrational $r$ such that $a + br$ is a rational number. Can you use this assumption to derive some nonsensical result, and thus show it must be false? – rubikscube09 Oct 31 '17 at 17:22
  • Last line is irrelevent and utterly false. $b$ is rational, not an integer. There is no integer $k$ so that $\frac 12 = k + 1$, is there? – fleablood Oct 31 '17 at 17:23
  • Use the definition of rational. $a = \frac nm; n, m \in \mathbb Z; m \ne 0$ and $b = \frac rs; r,s \in \mathbb Z; r\ne 0, s\ne 0$. – fleablood Oct 31 '17 at 17:25
  • @fleablood wouldn't i need two more integer variables to set a+br equal to as well? –  Oct 31 '17 at 17:28
  • If $a + br$ is rational, yes, you would have two more integers. Do that and solve for $r$ is it possible for $r$ to be irrational when $a + br$ is rational. Is $r = \frac {(a + br) - a}{b}$ a rational if $a+br, b, a$ is rational? – fleablood Oct 31 '17 at 17:32
  • BTW if $b = 0$ then $a + br = a$ which is, of course, rational. That is the only reason for the condition $b \ne 0$. – fleablood Oct 31 '17 at 17:39
  • "wouldn't i need two more integer variables to set a+br equal to as well?" I was leaving that for you to figure out. – fleablood Oct 31 '17 at 17:40

3 Answers3

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Let's assume that $a+br$ is rational. On that premise, then there would exist an integer $c$ and nonzero integer $d$ such that $a+br = c/d$.

Then I'd manipulate the equation to lead to $r$ being expressed as the ratio of two integers, which is a contradiction.

Can you take it from here?

John
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What do you know about rational numbers? They can be written as a fraction. So your first step would be to assume $a+br$ is rational with $r$ being irrational, and then show that there is a contradiction. It would go something like

$$a+br = \frac{A}{B}$$ $$r = \frac{\frac{A}{B}-a}{b},$$ I trust you can take it from here.

ultrainstinct
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Use the definition of rational numbers.

$a = \frac nm$ for some integers $n,m$ where $m\ne 0$.

$b = \frac st$ for some integers $s,t$ where $t \ne 0$.

Since $b \ne 0$, $s = 0$.

$r \ne uw$ for any possible integers $uv$.

Prove that $a + br = \frac nm + \frac str \ne \frac pq$ for any possible integers $p, q$.

Hint: Do a proof by contradiction and solve for $r$.

..... or .......

Proposition: If $j,k$ then $j*k$, $\frac jk$ (assuming $k \ne 0$, and $j\pm k$ are rational.

Pf: That's for you to figure out. (You probably have already had that as a theorem.)

If $a+ br = q$ then $r = \frac {q-a}b$. If $q$ is rational then $q -a$ is rational. If $q - a$ is rational then $\frac {q-a}b = r $ is rational. But $r$ is not rational. So $q$ being rational is impossible.

fleablood
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