Is the square root of 7 irrational?

Steps to Solve

1. Definition of Irrational Numbers

Irrational numbers are numbers that cannot be expressed as a fraction of two integers. In other words, if a number cannot be written in the form $ \frac{p}{q} $ where $ p $ and $ q $ are integers and $ q \neq 0 $, it is classified as irrational.

2. Assume Square Root of 7 is Rational

Let's assume, for the sake of contradiction, that $ \sqrt{7} $ is a rational number. This means we can write it as $ \frac{p}{q} $, where $ p $ and $ q $ are integers with no common factors (i.e., they are coprime).

3. Square Both Sides

If $ \sqrt{7} = \frac{p}{q} $, then squaring both sides gives us:

$$ 7 = \frac{p^2}{q^2} $$

By rearranging this equation, we can express it as:

$$ p^2 = 7q^2 $$

4. Analyze the Equation

The equation $ p^2 = 7q^2 $ suggests that $ p^2 $ is a multiple of 7. Since 7 is a prime number, this implies that $ p $ must also be divisible by 7. Let's denote $ p $ as $ p = 7k $ for some integer $ k $.

5. Substitute Back into the Equation

Substituting $ p $ back into the equation gives:

$$ (7k)^2 = 7q^2 $$

This expands to:

$$ 49k^2 = 7q^2 $$

6. Simplify the Equation

Now divide both sides by 7:

$$ 7k^2 = q^2 $$

This tells us that $ q^2 $ is also a multiple of 7, which subsequently implies that $ q $ must also be divisible by 7.

7. Conclusion

Since both $ p $ and $ q $ are divisible by 7, this contradicts our original assumption that $ p $ and $ q $ have no common factors. Therefore, our assumption that $ \sqrt{7} $ is rational must be false, which means:

$$ \sqrt{7} \text{ is irrational.} $$