If \(\rm\frac{{61}}{{19}}{\rm{}} = {\rm{}}3{\rm{\;}} + {\rm{\;}}\frac{1}{{x\; + \;\frac{1}{{y\; + \;\frac{1}{z}}}}}\) where x, y and z are natural numbers, then what is z equal to?
3
Solving Continued Fraction Problems
This problem involves a continued fraction. We are given an equation where a fraction \(\frac{61}{19}\) is expressed in a specific continued fraction form, and we need to find the value of the natural number 'z'.
Understanding Continued Fractions
A continued fraction is an expression obtained through an iterative process of representing a number as a sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on.
The general form given is: \[\frac{{61}}{{19}}{\rm{}} = {\rm{}}3{\rm{\;}} + {\rm{\;}}\frac{1}{{x\; + \;\frac{1}{{y\; + \;\frac{1}{z}}}}}\] where x, y, and z are natural numbers (positive integers: 1, 2, 3, ...).
Step-by-Step Solution
We need to express the fraction \(\frac{61}{19}\) in the given continued fraction form by repeatedly separating the integer part and taking the reciprocal of the fractional part.
- First, separate the integer part of \(\frac{61}{19}\): \[\frac{61}{19} = 3 + \frac{4}{19}\] Comparing this with the given form \(3{\rm{\;}} + {\rm{\;}}\frac{1}{{x\; + \;\frac{1}{{y\; + \;\frac{1}{z}}}}}\), we can see that the remaining part \(\frac{4}{19}\) must be equal to \(\frac{1}{{x\; + \;\frac{1}{{y\; + \;\frac{1}{z}}}}}\).
- Now, take the reciprocal of \(\frac{4}{19}\) to match the structure: \[\frac{4}{19} = \frac{1}{\frac{19}{4}}\] So, we have \(\frac{19}{4} = x\; + \;\frac{1}{{y\; + \;\frac{1}{z}}}\).
- Next, separate the integer part of \(\frac{19}{4}\): \[\frac{19}{4} = 4 + \frac{3}{4}\] Comparing this with \(x\; + \;\frac{1}{{y\; + \;\frac{1}{z}}}\), we can identify \(x = 4\). The remaining part \(\frac{3}{4}\) must be equal to \(\frac{1}{{y\; + \;\frac{1}{z}}}\).
- Take the reciprocal of \(\frac{3}{4}\) to match the structure: \[\frac{3}{4} = \frac{1}{\frac{4}{3}}\] So, we have \(\frac{4}{3} = y\; + \;\frac{1}{z}\).
- Finally, separate the integer part of \(\frac{4}{3}\): \[\frac{4}{3} = 1 + \frac{1}{3}\] Comparing this with \(y\; + \;\frac{1}{z}\), we can identify \(y = 1\). The remaining part \(\frac{1}{3}\) must be equal to \(\frac{1}{z}\).
- From \(\frac{1}{3} = \frac{1}{z}\), we can conclude that \(z = 3\).
We are given that x, y, and z are natural numbers. Our calculated values are x=4, y=1, and z=3, which are all natural numbers. Therefore, the solution is valid.
The value of z is 3.
Revision Table: Continued Fraction Expansion Steps
| Step | Calculation | Comparison | Identified Value |
|---|---|---|---|
| 1 | \(\frac{61}{19} = 3 + \frac{4}{19}\) | \(3 + \frac{4}{19} = 3 + \frac{1}{x + \dots}\) | |
| 2 | \(\frac{4}{19} = \frac{1}{\frac{19}{4}}\) | \(\frac{1}{\frac{19}{4}} = \frac{1}{x + \frac{1}{y + \frac{1}{z}}}\) implies \(\frac{19}{4} = x + \frac{1}{y + \frac{1}{z}}\) | |
| 3 | \(\frac{19}{4} = 4 + \frac{3}{4}\) | \(4 + \frac{3}{4} = x + \frac{1}{y + \frac{1}{z}}\) | \(x = 4\) |
| 4 | \(\frac{3}{4} = \frac{1}{\frac{4}{3}}\) | \(\frac{1}{\frac{4}{3}} = \frac{1}{y + \frac{1}{z}}\) implies \(\frac{4}{3} = y + \frac{1}{z}\) | |
| 5 | \(\frac{4}{3} = 1 + \frac{1}{3}\) | \(1 + \frac{1}{3} = y + \frac{1}{z}\) | \(y = 1\) |
| 6 | \(\frac{1}{3} = \frac{1}{z}\) | \(\frac{1}{3} = \frac{1}{z}\) | \(z = 3\) |
Additional Information on Continued Fractions and Natural Numbers
Continued fractions are a powerful tool in number theory and approximation theory. Every real number can be represented as a continued fraction.
- For rational numbers (like \(\frac{61}{19}\)), the continued fraction is finite.
- For irrational numbers (like \(\sqrt{2}\) or \(\pi\)), the continued fraction is infinite.
- The values \(x, y, z, \dots\) in a simple continued fraction \(a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \dots}}\) are typically integers, with \(a_1, a_2, \dots\) being positive integers (natural numbers). In this problem, x, y, and z are specifically stated to be natural numbers.
- The method used here is essentially the Euclidean algorithm applied to finding the greatest common divisor (GCD), but expressed in terms of fractions.
Natural numbers are the counting numbers: 1, 2, 3, 4, 5, ... They are a fundamental concept in mathematics.
The problem leverages the unique representation of a rational number as a finite continued fraction.
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