Is the Square Root of 27 a Rational Number?
Understanding the nature of the square root of 27 as a rational or irrational number is crucial in the realm of mathematics. In this article, we will explore the concept, provide the necessary proofs, and explain why the square root of 27 is not a rational number.
Definition of Rational and Irrational Numbers
A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, where p is the numerator and q is the non-zero denominator. On the contrary, an irrational number is a real number that cannot be expressed as a simple fraction and has a non-terminating, non-repeating decimal expansion.
Exploring the Square Root of 27
The square root of 27, denoted as sqrt{27}, can be simplified as follows:
sqrt{27} sqrt{9 times 3} sqrt{9} times sqrt{3} 3sqrt{3}
Since sqrt{3} is an irrational number, multiplying it by 3 (which is a rational number) still results in an irrational number. Therefore, 3sqrt{3} is also irrational, and consequently, the square root of 27 is not a rational number.
Alternative Proof Using Rational Number Definition
To further confirm that sqrt{27} is not a rational number, we can use the definition of a rational number. A number is rational if it can be expressed in the form p/q where p and q are integers and q is not zero. For sqrt{27}, if it were rational, it would have to be expressible as:
sqrt{27} a/b, where a and b are integers, and b ≠ 0
Squaring both sides, we get:
27 a^2/b^2
Multiplying both sides by b^2, we get:
27b^2 a^2
This implies that a^2 is a multiple of 27. Hence, a must be a multiple of 3. Let a 3c where c is an integer. Substituting a back into the equation, we get:
(3c)^2 27b^2
9c^2 27b^2
3c^2 9b^2
c^2 3b^2
This implies that c^2 is a multiple of 3. Therefore, c must also be a multiple of 3. This creates an infinite descent, indicating that our initial assumption that sqrt{27} is rational leads to a contradiction. Hence, sqrt{27} is not a rational number.
Proving the Irrationality of sqrt{3}
Given that sqrt{27} 3sqrt{3}, it suffices to prove that sqrt{3} is irrational. There are many proofs available that demonstrate the irrationality of sqrt{3}. One common method is the proof by contradiction approach:
Assume sqrt{3} is rational. Then, there exist integers p and q with no common factors, and sqrt{3} p/q. Squaring both sides, we get 3 p^2/q^2. Multiplying both sides by q^2, we obtain 3q^2 p^2. This implies that p^2 is a multiple of 3, so p must be a multiple of 3. Let p 3k, where k is an integer. Substituting p back, we get 3q^2 (3k)^2. Therefore, 3q^2 9k^2. This implies that q^2 is a multiple of 3, so q must be a multiple of 3. This contradicts the initial assumption that p and q have no common factors. Hence, our assumption that sqrt{3} is rational is false, and sqrt{3} is irrational.Therefore, sqrt{27} 3sqrt{3} is also irrational.
Conclusion
In conclusion, the square root of 27 is not a rational number. It is an irrational number, just like the square root of 3.