Is The Square Root Of 8 A Rational Number?
Table of Contents :
To determine whether the square root of 8 is a rational number, we first need to clarify what rational numbers are. A rational number is defined as a number that can be expressed as the fraction of two integers ( \frac{a}{b} ), where ( a ) and ( b ) are integers, and ( b \neq 0 ). This means that rational numbers can be whole numbers, fractions, or decimals that terminate or repeat. In contrast, irrational numbers cannot be expressed as simple fractions and their decimal expansions are non-repeating and non-terminating.
Understanding the Square Root of 8
To find the square root of 8, we can express it in a more manageable form.
[ \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2} ]
Now, the key part here is to evaluate the term ( \sqrt{2} ).
Is ( \sqrt{2} ) Rational?
It's widely established in mathematics that ( \sqrt{2} ) is an irrational number. This can be shown through a proof by contradiction:
-
Assume that ( \sqrt{2} ) is rational. This means we can write it as ( \frac{p}{q} ), where ( p ) and ( q ) are coprime integers (having no common factors other than 1).
-
Then we have:
[ \sqrt{2} = \frac{p}{q} \Rightarrow 2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2 ]
-
This implies that ( p^2 ) is even (since it is equal to ( 2q^2 )).
-
Therefore, ( p ) must also be even (because the square of an odd number is odd). Thus, we can say ( p = 2k ) for some integer ( k ).
-
Substituting back, we have:
[ (2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow 2k^2 = q^2 ]
-
This implies that ( q^2 ) is even, leading to the conclusion that ( q ) must also be even.
-
This leads to a contradiction since both ( p ) and ( q ) cannot be even (they are coprime).
Thus, since ( \sqrt{2} ) cannot be expressed as a fraction of two integers, it is confirmed that ( \sqrt{2} ) is irrational.
Conclusion on ( \sqrt{8} )
Since ( \sqrt{8} ) can be rewritten as ( 2\sqrt{2} ), and since we have established that ( \sqrt{2} ) is irrational, it follows that:
[ \sqrt{8} = 2\sqrt{2} ]
is also irrational.
Summary Table
To summarize, here’s a comparison:
<table> <tr> <th>Type</th> <th>Examples</th> </tr> <tr> <td>Rational Numbers</td> <td>1, -3/4, 0.75, 2.5 (terminating and repeating)</td> </tr> <tr> <td>Irrational Numbers</td> <td>√2, √3, π, e (non-terminating and non-repeating)</td> </tr> </table>
Important Note
It is essential to remember that not all square roots are rational. Specifically, the square roots of non-perfect squares (like 2, 3, 5, 6, etc.) result in irrational numbers. Therefore, when dealing with square roots, particularly in mathematical explorations or applications, it's crucial to evaluate whether the number under the square root is a perfect square or not.
In conclusion, the square root of 8 is indeed an irrational number because it can be expressed as ( 2\sqrt{2} ), where ( \sqrt{2} ) itself is confirmed to be irrational. Thus, the initial question, "Is the square root of 8 a rational number?" can be definitively answered with a resounding no.