I need help with a riddle my Algebra teacher gave us?
Why did the square root of 9 refuse to discuss anything with his brothers cubed root of 9, 1 ninth and the square root of negative 9?
ONE WAS IRRATIONAL
ONE KEPT ON REPEATING HIMSELF
ONE WAS UNREAL
To solve this riddle, we need to understand the properties and relationships of square roots and cube roots. Let's break it down step by step.
1. Start by finding the square root of 9: √9 = 3. The square root of a number gives us a value that, when squared, equals the original number.
2. Next, find the cube root of 9: ∛9 ≈ 2.08. The cube root of a number gives us a value that, when cubed, equals the original number.
3. Now, let's consider the square root of -9: √(-9). This is where things become interesting. The square root of a negative number (√(-9)) does not yield a real number. It is denoted as an imaginary number, specifically, √(-1) multiplied by the square root of the absolute value of -9, which simplifies to 3i.
With this understanding, let's look at the riddle again:
Why did the square root of 9 refuse to discuss anything with his brothers cube root of 9, 1 ninth, and the square root of negative 9?
The reason the square root of 9 refuses to discuss with the other three expressions is that the square root of negative 9 (√(-9)) is an imaginary number (3i), while the remaining three expressions are real numbers (3, 1/9, and approximately 2.08).
Therefore, the square root of 9 refuses to discuss with the cube root of 9, 1/9, and the square root of negative 9 because they belong to different number systems (real and imaginary numbers).
Brother cubed root of 9 is irrational.
Brother 1/9 repeats himself.
Brother square root of -9 is not real (he is imaginary).