Getting to the root

Last time, we have established the existence of irrational numbers without any mention of square roots. Why now is e. g $\sqrt{2}$ among them? The standard proof overuses variables and cannot be easily generalized. Here I present an purely arithmetical proof that establishes a simple condition for the irrationality of any root of any rational (i. e. of the form $\sqrt[n]{\frac pq}$, $p,q,n\in\mathbb N$).

The overarching idea is to work through a fair selection of arithmetical examples until the general method (in this case: of finding a square root) becomes visible. Then, present examples for which the method obviously fails, and a general criterion can be found by comparing and contrasting the examples' essential and contiguous elements. A good selection of arithmetical examples exhausts all thinkable possibilities and can thus serve as basis for a discussion of the general case. Such a proof "by prototyping" may lack in rigor and formality, but gains in clarity and constructiveness.

Easing in

Let's start by finding the root of a few square numbers:

$\sqrt{49}=\sqrt{7^2}=7, \sqrt{121}=\sqrt{11^2}=11, \sqrt{81}=\sqrt{9^2}=9.$

Easy enough. What about larger numbers, e. g. $\sqrt{196}$? We decompose it into prime factors and identify double factors:

$196 = 2\times 2\times 7\times 7 = (2\times 7)\times (2\times 7)=14\times 14=14^2 \Longrightarrow \sqrt{196}=\sqrt{14^2}=14.$

We can even flex our muscles with the laws of exponents:

$576 = 2^6\times 3^2 = (2^3\times 3)^2 \Longrightarrow \sqrt{576}=2^3\times 3 = 24$

$\sqrt{5625}=\sqrt{3^2\times 5^4}=\sqrt{(3\times 5^2)^2}=3\times 5^2=75$

$\sqrt{9801}=\sqrt{3^4\times 11^2}=3^2\times 11=99.$

These three examples show (in varying notation) the general method:

  • decompose the radicand into prime factors
  • pair up the factors (i. e. write the product as a square)
  • collect the single prime factors into the result.

The sky's the limit:

 
 

The factors don't necessarily have to be prime, as long as they can be paired without leftovers:

$\sqrt{100^{100}}=\sqrt{(10^{50})^2}=10^{50}.$

Sure, but…

All this is fine and dandy as long as the radicand has been specially prepared by the teacher. What about:

$\sqrt{99}=\sqrt{3^2\times 11}$?

The prime factorization tells us the ugly truth: the radicand is not a square number, so the root cannot possibly be an integer. Our calculator confirms this, but can only spew out decimal digits, not an exact value. At this point, the best we can do is partial reduction: extracting whatever double factors there are and keeping the leftovers under the root:

$\sqrt{3^2\times 11}=\sqrt{3^2}\times \sqrt{11}=3\sqrt{11}.$

The root is now at least "simpler", but an exact value is still lacking. At least we have reduced the problem to a "smaller" (yet still infinite) class of square roots. Once we know what $\sqrt{11}$ is, we also have a value for $\sqrt{99}$.

in any case, we can formulate the condition for the factor method to succeed as a theorem:

Theorem. The square root of a positive integer $N$ is itself an integer precisely when all prime factors of $N$ occur in pairs.
Proof. The square $R^2$ of any positive integer $R$ contains every prime factor of $R$ twice (counted with their multiplicities). An integer $N$ with an unpaired prime factor thus cannot be the square of some integer $R$.

Hope springs Rational

What does it mean to "know" what $\sqrt{11}$ "is"? Bear in mind that for a student who first encounters the root symbol, $\sqrt{11}$ is not simply a number but a problem awaiting an answer in decimal or fractional notation. Make sure your students view fractions as an equally valid, if not better representation than decimals. Then the shortcomings of the calculator answer become apparent. In order to strengthen this motivation, you can ask calculator-clingy students to e. g.:

  • check the numerical result by squaring: a cheap model will return $1.41421356^2\simeq 1.99999999$
  • calculate $\sqrt{1000000001}\simeq 10000.00005$: here the available significant digits are taken up by zeroes, and the "interesting" tail is rounded off

In any case, whether by persuasion or habit, instill in your students the aim to find an exact rational answer for the square roots that don't resolve to an integer.

Rational Radicands

This might seem to be the perfect moment to present a proof that e. g. $\sqrt{2}$ is irrational. I recommend that you resist this urge, and instead turn your students' attention to extending the factor method to rational radicands, leaving the problem of unresolved integer radicands for now. Let them work out:

$\sqrt{\frac{16}{25}} = ?$

$\sqrt{\frac{289}{81}} = ?$

$\sqrt{\frac{125}{245}} = ?$

$\sqrt{0.0625} = ?$

These examples illustrate that the factor method works just as well for fractions (some reducing and expanding may apply). Once the numerator and denominator are both written as products of paired factors, the root can be resolved:

 
 

The denominator and numerator decompose separately, and therefore we can extend the condition for integers:

Theorem. The square root of a positive rational $\frac pq$ (reduced form) is itself a rational precisely when all prime factors of $p$ occur in pairs, and the same condition for $q$.
Proof. The square $(\frac nm)^2$ of any positive rational $\frac nm$ contains every prime factor of $\frac nm$ twice (counted with their multiplicities). An integer $\frac pq$ with an unpaired prime factor in $p$ or $q$ thus cannot be the square of some rational $\frac nm$.

This now answers the open question of the unresolved square roots:

Corollary. The roots of the non-square integers (i. e. all naturals except 0, 1, 4, 9, 16, …) are all irrational. Therefore
$\sqrt{2}\simeq 1.4142135623\dots$
$\sqrt{3}\simeq 1.7320508075\dots$
$\sqrt{5}\simeq 2.236067977\dots$
$\sqrt{6}\simeq 2.4494897427\dots$
etc.
are all infinitely long decimals that never repeat.

This formulation is much more general than the usual restricted result $\sqrt 2\notin\mathbb Q$ (and maybe, as an exercise, $\sqrt 3\notin\mathbb Q$), and still just a special case of the preceding theorem. Yet both can be shown on arithmetical examples alone.

Higher-order Roots

Finally, the extension to cubic roots and general roots is straight-forward:

Theorem. The $n$-th root $\sqrt[n]{\frac pq}$ of a positive rational $\frac pq$ (in reduced form, $p,q,n\in\mathbb N$) is rational precisely if the prime factors of $p$ can be grouped into sets of $n$ identical factors each, with no factors left, and the same condition for $q$.

Concluding Remarks

The irrationality of roots is a topic where algebraic notation obscures more than it clarifies, and thus I deem the present, arithmetical approach more valuable for students. It is much closer to the operational day-by-day work, while still providing valid reasons for general theorems. In a later post, I will expand more generally on the benefits and possible critiques of this kind of proof "by prototypes".