Prove: The Square Root of a Prime Number is Irrational.

In our previous lesson, we proved by contradiction that the square root of 2 is irrational. This time, we are going to prove a more general and interesting fact. We will also use the proof by contradiction to prove this theorem.

That is, let p be a prime number then prove that \sqrt p is irrational.

Theorem: Suppose P is a prime number. The square root of P is irrational.

But first, let’s define a prime number. A prime number is a positive integer greater than 1 that has exactly two positive integer divisors: namely, 1 and itself.

To see it for yourself, below is the list of the first ten (10) prime numbers. Notice that each of them is only divisible by \large{1} and itself. Just a side-note, the number 2 is the only EVEN prime number.

  • 2
  • 3
  • 5
  • 7
  • 11
  • 13
  • 17
  • 19
  • 23
  • 29

To prove this theorem, we will use the method of Proof by Contradiction. We will assume the negation (or opposite) of the original statement to be true. That is, let \color{red}p be a prime number and \sqrt {\color{red}p} is a rational number. Now, the line of thought is to prove that \sqrt {\color{red}p} is rational. However, we expect a contradiction such that we discard the assumption, and therefore claim that the original statement must be true, which in this case, the square root of a prime number is irrational.

Since we assume that \sqrt p is rational, it means there exists two positive integers a and b but b \ne 0 that we can express as a ratio like the one below.

sqrt(p)=a/b

An essential part of this proof is to further assume that the greatest common divisor of a and b is 1. We can write it symbolically in math as \gcd \left( {a,b} \right) = 1.

This equation is asking to be squared on both sides, and see what we can make sense of it after doing so.

\left( {\sqrt p } \right)^2 ={ \Large{{\left( {{a \over b}} \right)^2}}}

p=a^2/b^2

Multiply both sides of the equation by b^2 to get rid of the denominator. This is the result.

{b^2}\,p = {a^2}

I want to move around the equation so that it is much easier to understand what it is trying to say. I hope you agree that the equation above is exactly the same as the one below.

a^2=pb^2

Since a is a positive integer greater than 1 then you can express it as a product of unique prime numbers with even or odd powers. However, by raising a to the power of 2, a^2 must have prime factorizations wherein each unique prime number will have an even exponent.

Let’s have an example to amplify what I meant above. Suppose a = 3,780. Breaking it down as a product of prime numbers, we get a = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5 \cdot 7. We can condense the prime factorization by rewriting it as a = {2^2} \cdot {3^3} \cdot 5 \cdot 7. Notice how we removed duplicates of prime numbers by expressing it as factors of primes with each unique prime having the appropriate power. The exponent tells how many times the prime number appears in the prime factorization.

prime factorization of the integer 3,780

Notice that in the prime factorization of integer \color{blue}\large{a}, the prime numbers can either have an odd or even exponent. This is an important observation that we will take advantage of later.

The next step is to square the integer \color{blue}\large{a}, thus we have \color{blue}\large{a^2}.

Apply the Power of a Power Rule of Exponent. We will use this property to square the integer \color{blue}\large{a}. In a nutshell, this exponent rule will allow us to distribute the main exponent 2 to the exponents of the unique prime numbers.

a^2=2^4 times 3^6 times r^2 times 7^2

After squaring the integer \color{blue}\large{a}, the exponents of the unique prime factors of \color{blue}\large{a} are now ALL even numbers. It is the case since the product of 2 and any integer will always be an even number. This is the most important observation that we can take from this step. We will definitely revisit this result.

Why did we spend some time prime factorizing the integer above? The reason is to demonstrate or illustrate by example the Fundamental Theorem of Arithmetic which is central to the proof.


The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that for every integer n greater than one, n > 1, we can express it as a prime number or product of prime numbers. The theorem further asserts that each integer has a unique prime factorization thus it has a distinct combination or mix of prime factors. In other words, the prime factorization of an integer is so unique because each prime factor always appears in the same amount or quantity thereby the arrangement doesn’t matter.

In order to have uniformity in our application of the fundamental theorem of arithmetic, we have to agree that we write the prime factors of an integer in ascending order. This way we can clearly see without any doubt that each integer’s prime factorization (greater than 1) is clearly unique.

Examples:

100 = 2 \cdot 2 \cdot 5 \cdot 5 = {2^2} \cdot {5^2}
126 = 2 \cdot 3 \cdot 3 \cdot 7 = 2 \cdot {3^2} \cdot 7
5,070 = 2 \cdot 3 \cdot 5 \cdot 13 \cdot 13 = 2 \cdot 3 \cdot 5 \cdot {13^2}
12,375 = 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 11 = {3^2} \cdot {5^3} \cdot 11

The next natural step is to generalize the factorization of any integer greater than 1 using the fundamental theorem of arithmetic.

Recall that in Equations #1, #2, and #3, it is clear that we must show how to factorize integers a and b in general form. And more importantly, we will need to examine what happens  to the prime factorizations of a and b when we square both of them, that is a^2 and b^2.

I’m currently working on this lesson. This is not finished yet. Thank you for your patience!