Proof that the Solution of 3^x = 8 is Irrational

In this post, we discuss another example of proof by contradiction. We are going to prove that the solution of the equation 3^x = 8 is irrational.

Theorem

The solution to the equation $latex 3^x = 8$ is irrational.

Proof

In proof by contradiction, we assume the opposite of the theorem and then find a contradiction somewhere in the proof.

Let us assume the opposite of the theorem. We assume that the solution to the equation $latex 3^x = 8$ is rational.

Now assuming that that the solution to the equation above is rational, then $latex x$ is a rational number. This means that $latex x =\frac{p}{q}$, where $latex p$ and $latex q$ are integers and $latex q$ not equal to $latex 0$ (definition of rational numbers). Therefore,

$latex 3^{\frac{p}{q}} = 8$

Raising both sides to $latex q$, we have $latex 3^p= 8^q$.

Notice that this is a contradiction since $latex 3^p$ is odd and $latex 8^q$ is even and therefore the two numbers cannot be equal. In addition, $latex 3^p = 2^{3q}$ contradicts the Unique Factorization Theorem or the Fundamental Theorem of Arithmetic.

Therefore, our assumption is false and the theorem that $latex x$ is irrational is true.

 

Proof that Square Root of Three is Irrational

Introduction

In The Intuitive Proof of the Infinitude of Primes, I showed you a proof strategy called proof by contradiction.  In this post, we use this strategy to prove that $latex \sqrt{3}$ is irrational.  In proof by contradiction, if the statement P is true, you have to assume the contrary, and then find a contradiction somewhere. Note that the proof in this post is very similar to the proof that $latex \sqrt{2}$ is irrational.

square root of three

Theorem:  $latex \sqrt{3}$ is irrational. Continue reading