Problem 45 - Algebra 2

Question

The golden ratio is \(\frac{1+\sqrt{5}}{2} .\) Find the difference between the golden ratio and its reciprocal.

Step-by-Step Solution

Answer

The difference between the golden ratio and its reciprocal is \(\sqrt{5}\).

Step 1: Find reciprocal of golden ratio

Firstly, find the reciprocal of the golden ratio. The reciprocal of a number is simply 1 divided by the number. So the reciprocal of the golden ratio \(\phi\), denoted as \(1/\phi\), will be \(2/(1 + \sqrt{5})\).

Step 2: Simplify the reciprocal

Simplify the reciprocal of the golden ratio. To do this, multiply the numerator and the denominator by the conjugate of the denominator, which is \(1 - \sqrt{5}\). This then gives us: \(1/\phi = (2 \times (1 - \sqrt{5})) / ((1+\sqrt{5}) \times (1-\sqrt{5})) = 2 - 2\sqrt{5}\).

Step 3: Find the difference

Now, subtract the reciprocal of the golden ratio from the golden ratio itself. This will give the difference between the two values : \(\phi - 1/\phi = (1 + \sqrt{5})/2 - (2 - 2\sqrt{5}) = \sqrt{5}\).

Key Concepts

Reciprocal

When we talk about the reciprocal of a number, we're referring to the basic idea of flipping it upside down. To find the reciprocal, you simply divide 1 by that number. For instance, the reciprocal of a whole number, like 7, would be \( \frac{1}{7} \). In our exercise, we are asked to find the reciprocal of the golden ratio, which is given as \( \frac{1 + \sqrt{5}}{2} \). This involves placing 1 over the golden ratio to get its reciprocal, represented as \( \frac{2}{1 + \sqrt{5}} \). It's a key step in many mathematical exercises and helps in understanding how numbers relate to each other in multiplication and division.

• The concept of a reciprocal is important for division and algebraic manipulations.
• Reciprocals are used to reverse the effects of multiplication. For instance, multiplying a number by its reciprocal always results in 1.

Simplification

Simplification refers to the process of making an expression easier to work with. In our exercise, the simplification process is applied to the reciprocal of the golden ratio. To simplify \( \frac{2}{1 + \sqrt{5}} \), we multiply both the numerator and the denominator by the conjugate of the denominator.
A conjugate in mathematics is the same algebraic expression but with the sign in the middle switched. So the conjugate of \( 1 + \sqrt{5} \) is \( 1 - \sqrt{5} \). By multiplying by this conjugate, the denominator becomes a rational number.
The simplification of the reciprocal thus becomes \( 2 \times (1 - \sqrt{5}) \) over \((1 + \sqrt{5})(1 - \sqrt{5})\), leading to a simpler form of \( 2 - 2\sqrt{5} \) after rationalizing the denominator.

• Using the conjugate is a common technique to remove square roots from the denominator, which aids in rationalization.
• Simplification often makes complex problems easier to handle by reducing the algebraic clutter.

Conjugate

In algebra, a conjugate is used to rationalize denominators or simplify the manipulation of expressions with roots. To find a conjugate, change the sign between two terms in a binomial. In the exercise above, the conjugate of \( 1 + \sqrt{5} \) is \( 1 - \sqrt{5} \). By multiplying by the conjugate, you essentially "remove" the square root from the denominator, turning it into a standard number.
This is because \((1 + \sqrt{5})(1 - \sqrt{5})\) results in a difference of squares: \(1 - 5\), which equals \(-4\). This transformation simplifies the problem into one that involves simpler arithmetic operations.

• Using a conjugate results in more manageable expressions in both algebra and calculus problems.
• The difference of squares formula, \(a^2 - b^2 = (a - b)(a + b)\), is key to understanding why conjugates are effective.