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5. For which values of \(n\) is \(\mathbb{Z}_{n}\) a field? Make a conjecture, test it, and try to prove it.
6. Say that an integer \(p\)_divides_ an integer \(n\) if \(n=pq\) for some integer \(q\), and say that \(p\) is _irreducible_ if whenever \(p\) divides \(nm\) then either \(p\) divides \(n\) or \(p\) divides \(m\). Show that \(\sqrt{p}\) is irrational for every irreducible integer \(p\).
### 1.3 Completeness and the Real Number System
What lies beyond the rational numbers? To answer this question, we begin with an observation.
**Observation 5**.: If \(n\) and \(m\) are odd, then \(nm\) is odd.
From this observation, we can draw an almost immediate conclusion: If a number \(n^{2}\) is a perfect square, it is either odd (if \(n\) is odd) or divisible by \(4\) (if \(n\) is even, then \(n=2m\) for one integer \(m\), so \(n^{2}=(2m)(2m)=4m^{2}\)). Then we have the following:
**Corollary 1**.: _Every even perfect square is divisible by \(4\)._
This simple observation has a profound consequence:
**Theorem 6**.: _There is no rational number \(\frac{p}{q}\) (where \(p,q\in\mathbb{Z}\) and \(q\neq 0\)) whose square is 2. That is, \(\sqrt{2}\) is irrational._
The argument we will use to prove Theorem 6 is called _proof by contradiction_. The strategy is to assume that there is a rational number whose square is \(2\) and then show that this assumption leads to a self-contradiction.
Proof
: Suppose \(\sqrt{2}=\frac{p}{q}\), where \(p\) and \(q\) are both integers in the lowest terms, i.e., with no common divisor greater than \(1\) (since if there were a common divisor, we could cancel it). Then \(2=\frac{p^{2}}{q^{2}}\), so \(p^{2}=2q^{2}\) and \(p^{2}\) is even. Then \(p^{2}\) is an even perfect square, and hence by Corollary 1, \(p^{2}\) is divisible by \(4\). Then \(p^{2}=2q^{2}=4x\) for some integer \(x\); then in particular, \(q^{2}=2x\). So \(q^{2}\) is even, which says \(q\) is even. But then \(p\) and \(q\) are both even, a contradiction since we chose \(\frac{p}{q}\) to be in the lowest terms. That is, _it is impossible_ to choose integers \(p\) and \(q\) with the greatest common divisor \(1\) so that \(\sqrt{2}=\frac{p}{q}\). Thus, there is no rational number \(\frac{p}{q}\) whose square is \(2\).
Hence, we see that there are numbers which are not rational, like \(\sqrt{2}\). A number \(x_{0}\) in \(\mathbb{R}\) is called _algebraic_ if it satisfies the equation
\[P(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}=0\]
for some polynomial \(P\) whose coefficients \(a_{0},a_{1},\ldots,a_{n}\) are integers, not all zero. Every rational number is algebraic and the number \(\sqrt{2}\) is also algebraic, since it satisfies the equation \(x^{2}-2=0\). Real numbers that are not algebraic are called _transcendental_. It is known, for example, that the numbers \(\pi\) and \(e\) are transcendental. For more on transcendental numbers, see Chapter 2 in [20]. In the following we will discuss the properties of real numbers in more detail.
### The Real Numbers
The set of _real numbers_, denoted \(\mathbb{R}\), contains the rational numbers \(\mathbb{Q}\) as a subset. The operations of addition and multiplication on \(\mathbb{Q}\) extend to all of \(\mathbb{R}\) in such away that every element of \(\mathbb{R}\) has an additive inverse and every nonzero element of \(\mathbb{R}\) has a multiplicative inverse. Real numbers also obey the _order axioms_ from \(O_{1}\) through \(O_{4}\) given for \(\mathbb{Q}\). In fact \(\mathbb{R}\) is an _ordered field_, which contains \(\mathbb{Q}\) as a subfield. We will look at the real number system in much more detail later, when we define the concept of “least upper bound.”
It can be shown that the real numbers are the only ordered field with the property known as _completeness_. Geometrically, the real numbers form a continuous line with no gaps or holes. Although field axioms are fundamental to the real numbers, by themselves they do not characterize \(\mathbb{R}\). In other words, the axioms of an ordered field are not enough to characterize \(\mathbb{R}\) uniquely. For example, the set of rational numbers \(\mathbb{Q}\) is a subfield of \(\mathbb{R}\) and also obeys these axioms. The one additional axiom that distinguishes \(\mathbb{R}\) from \(\mathbb{Q}\) is called the _completeness axiom_. The rational numbers by themselves are inadequate for analysis even though there are many rational numbers, in fact, a countably infinite number of them exist. Moreover, given any two rational numbers you can think of say, \(\dfrac{1}{100}\) and \(\dfrac{1}{101}\), one can find infinitely many rational numbers between these two. Yet it seems like there are not just enough of them. We have seen that the simple-looking equation \(x^{2}=2\) has no solution in the set of rational numbers \(\mathbb{Q}\). We showed \(\sqrt{2}\) is a real number, and it is the solution of the equation \(x^{2}=2\), but \(\sqrt{2}\notin\mathbb{Q}\). Thus the set of rational numbers contains “gaps.” We need a concept that clarifies what we mean by \(\mathbb{R}\), which does not contain any gaps or holes which also defines the difference between \(\mathbb{R}\) and \(\mathbb{Q}\).
### Axiom of Completeness:
Every nonempty set of real numbers that is bounded above has a least upper bound.
To understand the axiom of completeness, let us first define _bounded sets_ and _least upper bound_ of a set.
**Definition 7**.: A nonempty set \(A\subset\mathbb{R}\) is _bounded above_ if there exists a number \(M\) with \(a\leq M\) for all \(a\in A\). Then \(M\) is called an upper bound for \(A\). Similarly the set \(A\) is _bounded below_ if there exists a number \(m\) with \(a\geq m\) for all \(a\in A\). In this case \(m\) is a lower bound for \(A\).
Note that a given set can have many upper or lower bounds, for example, if \(A=(1,2)\), then A is bounded below by \(1\) or by \(-4,-36\), etc. \(A\) is bounded above by \(2\), or \(3,4,100\), etc. But if the set under consideration is \(A=(1,\infty)\), then \(A\) has lower bounds but has no upper bound.
**Definition 8**.: Let \(A\) be a nonempty subset of \(\mathbb{R}\) that is bounded above. The _supremum_ or _least upper bound_ of \(A\) is a number \(s\) such that
* \(s\) is an upper bound of \(A\).
* If \(b\) is another upper bound for \(A\), then \(s\leq b\).
Similarly, let \(A\) be a nonempty subset of \(\mathbb{R}\) that is bounded below. The _infimum_ or _greatest lower bound_ of \(A\) is a number \(s^{\prime}\) such that
* \(s^{\prime}\) is a lower bound of \(A\).
* If \(c\) is another lower bound for \(A\), then \(s^{\prime}\geq c\).
We write \(\sup A\) or \(\inf A\) to denote the supremum or the infimum of the set \(A\), respectively.
Note that the definition of infimum or supremum has two parts. For the least upper bound, the first part asserts that \(\sup A\) is an upper bound, and the second part states that \(\sup A\) must be the least one. It is not clear that supremum and infimum will always exist. However a finite set will always have a supremum or infimum. We just pick the biggest and the smallest element of the set, respectively. Although a set can have many upper bounds, it can only have one least upper bound. To see this assume the set \(A\) has two least upper bounds, say \(s_{1}\) and \(s_{2}\). Then from the definition of the least upper bound, we can assert \(s_{1}\leq s_{2}\) and \(s_{2}\leq s_{1}\); thus, \(s_{1}=s_{2}\) and the least upper bounds are unique (Figure 1.12).
**Example 18**.:
* Let \(A=\{-1,3,8,10\}\); then \(\inf A=-1\) and \(\sup A=10\). Both the infimum and the supremum belong to \(A\).
* Let \(B=\{1,3,5,7,\dots\}\). Then \(\inf B=1\), but \(\sup B\) does not exist.
* Let \(C=\left\{\frac{2\pi}{n}:\quad n\in\mathbb{N}\right\}\). Then \(\sup C=2\pi\) and \(\inf C=0\). The infimum does not belong to the set \(C\).
**Remark 5**.: Let \(A\) and \(B\) be nonempty bounded subsets of \(\mathbb{R}\); if
\[A+B:=\{a+b:a\in A\text{ and }b\in B\},\]
then it can be shown that \(\sup(A+B)=\sup(A)+\sup(B)\).
**Example 19**.: Let \(A\subset\mathbb{Q}\) be defined as
\[A=\{r\in\mathbb{Q}:\quad r^{2}<2\},\]
Figure 1.12: Sup A and \(\inf A\)
and suppose we are looking for the least upper bound for \(A\). Since \(\sqrt{2}=1.4142\ldots\), we might guess that \(b_{1}=\frac{142}{100}\) is an upper bound, but it is not the least one; for example, \(b_{2}=\frac{1415}{1000}\) is an upper bound smaller than the first one. Thus the question is can one find the smallest one? In the set of rational numbers we cannot.
We now give a “characterization” condition for the least upper bounds.
**Lemma 1** (Characterization of “\(\sup\)”).: _Assume \(s\in\mathbb{R}\) is an upper bound for a nonempty set \(A\subset\mathbb{R}\). Then \(s=\text{sup}A\) if and only if for every \(\epsilon>0\) there exists an element \(x\in A\) such that \(s-\epsilon : \(\Rightarrow\)) Assume \(s=\sup A\), and \(\epsilon>0\). We must produce an \(x\in A\) such that \(s-\epsilon (\(\Leftarrow\) suppose \(s\) satisfies the given condition. Let \(s^{*}\) be an upper bound of \(A\). According to the definition of \(\sup A\), we must show \(s\leq s^{*}\). Suppose \(s>s^{*}\), and then if we let \(\epsilon=s-s^{*}\), \(s^{*}=s-\epsilon\) and \(s^{*}\geq x\) for all \(x\in A\), which implies \(s-\epsilon\geq x\) or \(s\geq x+\epsilon\) and so our condition fails. Our assumption \(s>s^{*}\) is wrong, and therefore \(s\leq s^{*}\). Thus we verified both conditions in the definition of the least upper bound. **Remark 6**.: It is certainly the case that all of the above conclusions about \(\sup A\) have analogous versions for \(\inf A\). Furthermore the Axiom of Completeness can also be expressed as follows: Let \(A\) be a nonempty set in \(\mathbb{R}\) that has a lower bound. Then \(A\) has a greatest lower bound. The first application of the Axiom of Completeness is the Nested Interval Property, which expresses the fact that the real line contains no “gaps.” A collection of intervals \(I_{1},I_{2},I_{3},\dots\) is called _nested_ if \[I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq\dots.\] If the intervals \(I_{n}\) are open intervals: \[(0,1)\supseteq(0,1/2)\supseteq(0,1/3)\supseteq\dots,\] Figure 1.13: Characterization of “\(\sup\) then even though each interval contains infinitely many points, the intersection \(\bigcap_{n=1}^{\infty}I_{n}=\emptyset\). The intervals \(\{[n,\infty)\}_{n=1}^{\infty}\) are closed and nested but not bounded, and their intersection is empty. The story is different if we take both closed and bounded intervals, for example, if we consider the nested collection \[[0,1]\supseteq[0,1/2]\supseteq[0,1/3]\supseteq\dots,\] then their intersection has one point \(\bigcap_{n=1}^{\infty}I_{n}=\{0\}\) and thus is not empty.Proof
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