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Prove the following statement by proving its contra-positive:

If $ r $ is irrational, then $ r^{1/5} $ is irrational.

I am totally confused!

(1) How does proving the contra-positive prove the statement?

(2) The only way I know how to describe this is with the following try:

Let $ IR $ be the set of all irrational numbers. Let the predicate $ P(n) $ be such that $ n^{1/5} \in IR $

Thus,

$ \forall x \in IR, P(n) $.

Now what do I do? Am I on the right track?

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  • $\begingroup$ I guess you're following the MIT Mathematics in CS course. If so, there are readings provided in the course which will help you with the subsequent problems and assignments. These are complementary to the lectures and I didn't notice them at first either: ocw.mit.edu/courses/electrical-engineering-and-computer-science/… The problem above is explained in chapter 2 of the readings. $\endgroup$ – mallardz Sep 6 '14 at 8:28
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The first step is to write the contrapositive of the statement.

For a statement of the form, "$P$ implies $Q$", the contrapositive of that statement is, "not-$Q$ implies not-$P$". Therefore the two parts of your contrapositive statement should be "$r^{1/5}$ is not irrational" (or in other words, "$r^{1/5}$ is rational") and "$r$ is not irrational" (or in other words, "$r$ is rational".

The contrapositive of a statement is always equivalent to the original statement (they're either both true, or both false), so if you can prove one, you have proved the other.

You can prove the contrapositive of the problem as follows: assume $r^{1/5}$ is rational (the "not-$Q$" clause of the contrapositive), and show that $r$ is rational (the "not-$P$" clause of the contrapositive). It may help to clarify things if you let $x = r^{1/5}$ and write $r$ in terms of $x$, like so: $r = x^5$. So all you really need to prove is that if $x$ is rational, $x^5$ is also rational.

If $x$ is rational, then there are integers $m$ and $n$ ($n \ne 0)$ such that $x = m/n$. Express $x^5$ in terms of $m$ and $n$. Is the result a fraction with an integer in the numerator and an integer in the denominator (or can you make it into such a fraction)? If so, then $x^5$ is rational by the definition of "rational".

The statement "$r^{1/5}$ is irrational implies $r$ is irrational" is the converse of the statement you were to prove. The converse of a statement is not necessarily equivalent to the original statement, and proving the converse does not prove the original statement.

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  • $\begingroup$ This is a great answer, but I am not sure of the "proof" part. Where I get confused is the fact that if $ x $ is rational then $ x^{5} $ is rational seems obvious and I'm not sure if I need to prove something so incredibly obvious. Are we getting down to the axioms of numbers here, or the most basic facts of numbers? In other words, what is the actual proof step, and how do you know when you need to explicitly spell out the seemingly obvious? $\endgroup$ – Elegant Codeworks May 6 '14 at 7:55
  • $\begingroup$ If $x$ is rational, then there are integers $m$ and $n$ ($n \ne 0)$ such that $x = m/n$. Express $x^5$ in terms of $m$ and $n$. Is the result a fraction with an integer in the numerator and an integer in the denominator (or can you make it into such a fraction)? If so, then $x^5$ is rational by the definition of "rational". $\endgroup$ – David K May 6 '14 at 12:56
  • $\begingroup$ Actually that last comment of mine should have been part of the answer, so I've inserted it in the answer text. $\endgroup$ – David K May 6 '14 at 12:58
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The contrapositive of a statement is logically equivalent to the original statement. You can see this easily with truth tables.

Now $r$ is irrational implies $r^{1/5}$ is irrational is the original statement, so the contrapositive would be $r^{1/5}$ is rational implies $r$ is rational.

Additional hint: if $a, b$ are rational, then $ab$ is rational. This is known as closure.

Answer below

If $r^{1/5}$ is rational, then $r^{1/5} \cdot r^{1/5}\cdot r^{1/5}\cdot r^{1/5}\cdot r^{1/5} = r$ is rational.

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    $\begingroup$ Sorry, both numbers should be IRrational. I updated my question. I will give it a try from there, but can you post the answer so I can validate against it later? Perhaps at the bottom of your post. This is homework from MIT Open Courseware and I am "taking" it solely to learn for fun. $\endgroup$ – Elegant Codeworks May 6 '14 at 4:18
  • $\begingroup$ I have added the answer. Best of luck :) $\endgroup$ – Christopher Liu May 6 '14 at 4:21
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Let me answer your two questions separately:

  1. Any statement of the form "If $P$ then $Q$" is logically equivalent to its contrapositive, "If not $Q$ then not $P$." They mean exactly the same thing. But sometimes its easier or more natural to write a proof of one version than of the other.
  2. In this particular case, what you seem to be doing is interpreting the statement to be proved not as an implication of the form "If $P$ then $Q$" but as a single predicate with a universal quantifier ("$\forall x \in IR$, $P(x)$"). Instead, let $P$ be the statement "$r$ is irrational" and let $Q$ be the statement "$r^{1/5}$ is irrational". Then the statenent to be proved has the form "If $P$ then $Q$" and lends itself well to forming the contrapositive. Note that the negation of $P$ would be "$r$ is not irrational", or more simply "$r$ is rational".

Hope this helps!

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