Chapter 3: Problem 2
If \(a+b i\) is not a unit in \(J[i]\) prove that \(a^{2}+b^{2}>1\).
Short Answer
Expert verified
If \(a + bi\) is not a unit in \(\mathbb{Z}[i]\), then \(a^2 + b^2 > 1\).
Step by step solution
01
Understand Units in \(\mathbb{Z}[i]\)
In the ring of Gaussian integers, \(\mathbb{Z}[i]\), the units are elements that have a multiplicative inverse within \(\mathbb{Z}[i]\). The units in this set are \(\{1, -1, i, -i\}\). This means that if \(a + bi\) is not one of these units, it does not have a norm of 1.
02
Identify the Norm
For any Gaussian integer \(a + bi\), the norm is given by \(N(a + bi) = a^2 + b^2\). If \(a + bi\) is not a unit, then this norm must be greater than 1, as units have a norm of precisely 1.
03
Prove that \(a^2 + b^2 > 1\) for Non-units
Since \(a + bi\) is not a unit in \(\mathbb{Z}[i]\), its norm cannot equal 1. Given that the norm \(a^2 + b^2\) must be a positive integer, the next smallest value after 1 is 2. Thus, it follows that \(a^2 + b^2 > 1\).
04
Consider Possible Values of \(a\) and \(b\)
If \(a^2 + b^2 = 1\), then \(a, b\) must satisfy one of the pairs of values that make the sum of their squares equal to 1, namely \((1, 0), (-1, 0), (0, 1), (0, -1)\). These pairs correspond to the units \(1, -1, i, -i\). Thus, any \(a\) and \(b\) that do not form one of these units must satisfy \(a^2 + b^2 > 1\).
05
Conclusion
Since \(a + bi\) is not a unit in \(\mathbb{Z}[i]\), its norm \(N(a + bi) = a^2 + b^2\) is greater than 1. Therefore, we have proven that \(a^2 + b^2 > 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Units in \(\mathbb{Z}[i]\)
In mathematics, especially within the domain of Gaussian integers, units play a crucial role. A unit in a ring is an element that has a multiplicative inverse within the same ring. In the context of the Gaussian integers, \(\mathbb{Z}[i]\), these units are the numbers that can be multiplied by another Gaussian integer to result in 1.
The set of units in \(\mathbb{Z}[i]\) is well-defined and consists of just four elements: \( 1, -1, i, \) and \(-i\). These elements are special because their norms are exactly 1.
The set of units in \(\mathbb{Z}[i]\) is well-defined and consists of just four elements: \( 1, -1, i, \) and \(-i\). These elements are special because their norms are exactly 1.
- \(1\) and \(-1\) are real numbers and their multiplicative inverses are themselves in this set.
- Complex numbers \(i\) and \(-i\) also have multiplicative inverses within the Gaussian integers, satisfying \(i \times (-i) = 1\).
Norm of a Gaussian Integer
The concept of norm is central to understanding Gaussian integers. The norm of a Gaussian integer \(a+bi\) is calculated using the formula \(N(a+bi) = a^2 + b^2\). This norm is always a non-negative integer and provides insights into the properties of the Gaussian integer itself.
When we focus on units within \(\mathbb{Z}[i]\), such as \(1, -1, i, \text{ and } -i\), all have a norm equal to 1. This is because
When we focus on units within \(\mathbb{Z}[i]\), such as \(1, -1, i, \text{ and } -i\), all have a norm equal to 1. This is because
- For \(1\) or \(-1\), we compute \(1^2 + 0^2 = 1\).
- Likewise, for \(i\) or \(-i\), it's \(0^2 + 1^2 = 1\).
Non-unit Elements
Non-unit elements in \(\mathbb{Z}[i]\) are those Gaussian integers that are not \(1, -1, i, \) or \(-i\). For these elements, their properties differ significantly from units.
The norm of these non-unit elements is crucial for their classification. If \(a + bi\) is a non-unit, then \(a^2 + b^2\) will exceed 1. Let's consider why this matters:
The norm of these non-unit elements is crucial for their classification. If \(a + bi\) is a non-unit, then \(a^2 + b^2\) will exceed 1. Let's consider why this matters:
- Only the units can have a norm equal to 1. Any other element indicates a bigger footprint on the complex plane.
- When exploring pairs \(a, b\), valid integers that satisfy \(a^2 + b^2 = 1\) always trace back to the units \(1, -1, i, \) and \(-i\).
- Thus, all non-units inevitably result in \(a^2 + b^2 > 1\).