Problem 3 Show that $3 \sqrt{2}$ is an i... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 0: Problem 3

Show that $3 \sqrt{2}$ is an irrational number.

Short Answer

Expert verified

Assuming $3\sqrt{2}$ is rational and can be expressed as the simplified fraction $\frac{a}{b}$, where $a$ and $b$ are coprime integers, we derive an equation $18b^2 = a^2$. From this, we conclude that both $a$ and $b$ must be even numbers. However, this contradicts the initial assumption that $a$ and $b$ are coprime integers, as they share a common factor of $2$. Thus, $3\sqrt{2}$ must be an irrational number.

Step by step solution

Assume $3\sqrt{2}$ is Rational

Let's assume that $3\sqrt{2}$ is a rational number. That means it can be expressed as a simplified fraction $\frac{a}{b}$, where $a$ and $b$ are coprime integers: $3\sqrt{2} = \frac{a}{b}$ Now, let's manipulate this equation to eliminate the square root symbol.

Square Both Sides

To eliminate the square root symbol, we can square both sides of the equation: $(3\sqrt{2})^2 = (\frac{a}{b})^2$ Which simplifies to: $18 = \frac{a^2}{b^2}$

Re-arrange The Equation

To further simplify the equation, let's multiply both sides by $b^2$: $18b^2 = a^2$

Analyze Both Sides Of The Equation

The left side of the equation, $18b^2$, is obviously divisible by $2$ since $18$ is an even number and multiplying by $b^2$ won't change the fact that it is even. This means that $a^2$ on the right side of the equation is also divisible by $2$. Since $a^2$ is divisible by $2$, it implies that $a$ must also be an even number. So, let $a = 2m$ for some integer $m$.

Substitute $a$ and Simplify

Now, let's substitute $a = 2m$ back into the equation $18b^2 = a^2$: $18b^2 = (2m)^2$ Which simplifies to: $18b^2 = 4m^2$ Divide both sides by $2$: $9b^2 = 2m^2$

Analyze Both Sides Of The Equation Again

Now the right side of the equation, $2m^2$, is divisible by $2$. Since $9b^2$ is equal to $2m^2$, it implies that $9b^2$ is also divisible by $2$, and thus $b^2$ must be divisible by $2$. Since $b^2$ is divisible by $2$, it implies that $b$ must also be an even number. So, $a$ and $b$ are both even numbers.

Identify the Contradiction

We initially assumed that $a$ and $b$ are coprime integers. However, since both $a$ and $b$ are even numbers, they both have a common factor of $2$, which contradicts our initial assumption that they are coprime integers. Therefore, our initial assumption that $3\sqrt{2}$ is a rational number must be incorrect, and hence, $3\sqrt{2}$ is an irrational number.

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91影视

Show that \(3 \sqrt{2}\) is an irrational number.

Short Answer

Step by step solution

Assume \(3\sqrt{2}\) is Rational

Square Both Sides

Re-arrange The Equation

Analyze Both Sides Of The Equation

Substitute \(a\) and Simplify

Analyze Both Sides Of The Equation Again

Identify the Contradiction

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