Is this a valid proof of “a, b are rational, b ≠ 0, r is irrational, then a + br is irrational”

Theorem: If a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is irrational.

Proof: Assume that if a and b are rational numbers, b ≠ 0, and r is an irrational number, then a + br is rational.

By the definition of rational, we can substitute a and b with fractions where p, q, m, n are particular but arbitrary integers.

a = p/q   b = m/n

a + br = p/q + (m/n)r/1

= p/q + mr/n

= (pn + qmr) / qn

Since r is irrational, we know that both the numerator and the denominator cannot be rational numbers, which implies a + br is irrational, which contradicts the fact that a + br is rational. This contradiction shows the supposition is false, therefore the theorem is true.

• $(m/n)r\neq mr/nr$ if $m,r\neq 0$ and $(\sqrt{2}/3\sqrt{2})=1/3$ even if the numerator and denominator are irrationals. Let's try this. Suppose that $a+br=q$ where $q$ is rational. You have \begin{align*} a+br &= q\\ br&=q-a\\\end{align*} – MoebiusCorzer Jan 19 '16 at 18:51
• Your setup is fine, but the concluding argument is not convincing. Try solving for $r$ instead. – Dan Brumleve Jan 19 '16 at 18:56
• I believe r = (q - a) / b, which would make r a rational number (a contradiction), since it is the quotient of two rational numbers. Does that prove the proposition? – 123 Jan 19 '16 at 19:27
• Yes, you can also make it more explicit by expanding it to a ratio of two integers. – Dan Brumleve Jan 19 '16 at 19:40

One possibility is,since $$b\neq 0$$ and $$b$$ is rational we show first $$br$$ must be irrational.

This is done by Contradiction

Suppose $$br$$ is rational ,then since $$\frac{1}{b}$$ is rational, the product $$\frac{1}{b}(br)=r$$ is rational.(Products of rational numbers are rational)

But this is a contradiction because by Hypothesis $$r$$ is irrational.

Hence $$br$$ has to be irrational (First step proved)

Remains to show that since br is irrational as we have shown already that also the sum a+br is irrational.

If a is zero we have nothing more to show.

Otherwise we argue again by Contradiction and assume $$a+br$$ to be rational

Since $$-a$$ is rational(because a is rational) it follows that $$-a+(a+br)=br$$ is rational (sums of rational numbers are rational),which contradicts the irrationality of $$br$$ shown in the first step.

Hence $$a+br$$ must be irrational

(Second step proved)

q.e.d