Proof: $\sqrt{3}$ is an Irrational Number

Algebraic Method

Showing the proof that $\sqrt{3}$ is not rational involves a similar contradiction based approach to the proof that $\sqrt{2}$ is irrational.

The proof first starts by assuming that $\sqrt{3}$ is rational and can be written in the form of $\dfrac{p}{q}$ where $p$ and $q$ do not have any common factors. We square the expression to remove the square root sign.

The next step is to establish that $p$ must have 3 as a factor and thus can be replaced by $3p'$ . Simplifying this expression leads to the conclusion that $q$ must also have 3 as a factor. If both $p$ and $q$ have a common factor, it contradicts the early claim that we can write $\sqrt{3}$ as $\dfrac{p}{q}$ where $p$ and $q$ are integers in their lowest forms.

Best of luck in going through the step by step proof. You need to get all steps correct for the proof to count as successful.

Prove that $\sqrt{3}$ is an Irrational Number

We want to prove that $\sqrt{3}$ is irrational. What assumption should we begin with?

Suppose $\sqrt{3}$ is rational, so it can be written as $\dfrac{p}{q}$ .

What can we say about the integers $p$ and $q$ ?

Where $p, q \in \mathbb{Z}$ and $q \ne 0$ .

What assumption do we make about the form of $\dfrac{p}{q}$ ?

Assume that $\dfrac{p}{q}$ is in its lowest terms.

How do we express $\sqrt{3}$ as a rational number?

We can write $\sqrt{3} = \dfrac{p}{q}$

Simplify

$\Rightarrow 3 = \dfrac{p^2}{q^2}$

Simplify

$\Rightarrow p^2 = 3q^2$

What can we say about $p^2 = 3q^2$ ?

The L.H.S. must be divisible by 3 since R.H.S is a multiple of 3.

If $p^2$ is divisible by 3, then;

$\Rightarrow p$ is divisible by 3.

How do we represent $p$ to reflect divisibility by 3?

If $p$ is divisible by 3, we can say $p = 3p'$ where $p'$ is also an integer.

How do we write $p^2$ in terms of $p'$ ?

$\Rightarrow p^2 = 9p'^2$

Substitute this back into $p^2 = 3q^2$

Substitute $p^2$ with $9p'^2$ in $p^2 = 3q^2$

Substitution results in?

$\Rightarrow 9p'^2 = 3q^2$

Simplify

$\Rightarrow 3p'^2 = q^2$

What can we say about $q^2 = 3p'^2$ ?

Since L.H.S is divisible by 3, R.H.S must also be divisible by 3.

If $q^2$ is divisible by 3, then;

$\Rightarrow q$ is divisible by 3.

What contradiction do we reach?

Since both $p$ and $q$ are divisible by 3, they share a common factor of 3.

What conclusion do we draw from this contradiction?

This contradicts the assumption that $\dfrac{p}{q}$ is in lowest terms.

Final conclusion?

$\therefore$ Our initial assumption must be false. We cannot write $\sqrt{3}$ in the form of $\dfrac{p}{q}$ . Thus, $\sqrt{3}$ is irrational.

End of Lesson

Proof: 3\sqrt{3}3​ is an Irrational Number

Algebraic Method

Prove that 3\sqrt{3}3​ is an Irrational Number

Proof: 3\sqrt{3}3​ is an Irrational Number

Algebraic Method

Prove that 3\sqrt{3}3​ is an Irrational Number

Proof: $\sqrt{3}$ is an Irrational Number

Prove that $\sqrt{3}$ is an Irrational Number

Proof: $\sqrt{3}$ is an Irrational Number

Prove that $\sqrt{3}$ is an Irrational Number