Is the square root of 7 irrational?
Understand the Problem
The question is asking whether the square root of 7 is an irrational number. To determine this, we need to understand the definition of irrational numbers and check if the square root of 7 can be expressed as a quotient of two integers.
Answer
$ \sqrt{7} \text{ is irrational.} $
Answer for screen readers
The square root of 7 is an irrational number.
Steps to Solve
- 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.
- 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).
- 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:
$$ p^2 = 7q^2 $$
- 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 $.
- Substitute Back into the Equation
Substituting $ p $ back into the equation gives:
$$ (7k)^2 = 7q^2 $$
This expands to:
$$ 49k^2 = 7q^2 $$
- 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.
- 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:
$$ \sqrt{7} \text{ is irrational.} $$
The square root of 7 is an irrational number.
More Information
Interestingly, many square roots of numbers that are not perfect squares are irrational. The square root of prime numbers, like $ \sqrt{7} $, $ \sqrt{2} $, and $ \sqrt{3} $, all produce irrational numbers.
Tips
- Assuming that a number is rational just because it's a square root. Always check the nature of the number being squared and identify if it can be expressed as a fraction of two integers.
- Not following through with the proof by contradiction completely can create confusion regarding the nature of the numbers.