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There are some similar questions, but not exactly this one.

First:

As, Zero is a rational number. So, by using the counterexample $a=0$ and $b=x$ ($x$ being an irrational number) we get $ab = 0$. Thus it is not true that $ab$ necessarily has to be irrational.

Second:

However, I have learned a technique that I have used to proof similar but slightly different combination ($a$ is rational and $b$ is irrational, is $a+b$ irrational?) of questions.

With this technique (Proof by contradiction):

Given $a$ is rational and $b$ is irrational. Let, $ab$ be rational.

Suppose, $a= \frac{m}{n}$ and $ab= z =\frac{p}{q}$ , where $m,n,p,q$ are intgers and $n,q \neq 0$

$b=x$ where $x$ is an irrational number.

$$a \cdot b=z$$ $$\frac{m}{n} \cdot x=\frac{p}{q}$$ $$x=\frac{p}{q} \cdot \frac{n}{m}$$ Now, as $p,q,m,n$ are all integers, therefore $x$ is an integer, therefore rational, which contradicts our starting assumption that $x$ is an irrational number. Therefore, $ab$ is irrational.

Now, I think the First conclusion that $ab$ is not necessarily irrational is correct. The second proof is definitely wrong (I think), but I am new to proofs and don't know where it is wrong. What did I miss and not consider in the second approach?

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    $\begingroup$ The last equation of the second proof is invalid if $a = 0$ since then $m = 0$ $\endgroup$ – Kai Rüsch Mar 23 '17 at 3:05
  • $\begingroup$ Oh, Yes. I got too bogged down on rigid thinking. Then I suppose, if $a \neq 0$, i.e. for all other rational numbers except zero, won't the product of a rational and an irrational number be irrational ? $\endgroup$ – Rio1210 Mar 23 '17 at 3:10
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    $\begingroup$ To disprove that a statement is true for all values of something, it is perfectly fine to prove that it is not true for a single value. So showing that the statement fails for $a = 0$ is perfectly legitimate. $\endgroup$ – ConMan Mar 23 '17 at 3:16
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    $\begingroup$ Yes, the proof is correct if $a\neq 0$. $\endgroup$ – Friedrich Philipp Mar 23 '17 at 3:18
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    $\begingroup$ " The second proof is definitely wrong (I think)" I wouldn't say it is "definately" wrong. Id' say it is incomplete. $\frac mn x= \frac pq$ is correct. But the conclusion $x = \frac pq \frac nm$ needs to state $x = \frac pq \frac nm$ IF $m \ne 0$. So $ab$ is rational does lead to a contradiction UNLESS $a = 0$. In fact I'd say the proof is definitely correct except for one condition. $\endgroup$ – fleablood Mar 23 '17 at 6:14
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If $k = m/n$ is rational and $j = p/q\ne 0$ is rational, then $k/j = mq/np$ is rational (and if $j = 0$ then $k/j$ is not irrational; it is simply undefined and meaningless and not a number or anything at all).

So if $ab$ is rational. And $a$ is rational. And $a \ne 0$ then than $ab/a = b$ is rational.

So the only way. $a$ can rational and $b$ be irrational but somehow $ab $ is rational is if $ab/a$ is not defined. That only happens if $a$ is zero. That is the one and only counter example.

As for $a +b$.... Note if $k = m/n$ and $j = p/q$ are rational, then $k \pm j = \frac {mq \pm pn}{nq}$ is rational.

So if $a$ is rational and $a + b$ is rational, then $(b+a) - a = b$ must also be rational.

So if $a$ is rational and $b$ is irrational there is no way possible for $a + b$ to be rational. There are absolutely no counterexamples.

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Posting Kai Rüsch's answer-in-a-comment as an answer:

The last two equations of the second proof are:

$$\frac{m}{n} \cdot x = \frac{p}{q}$$

$$x = \frac{p}{q} \cdot \frac{n}{m}$$

This last equation is invalid if $a = 0$, since then $m = 0$.

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