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Chapter 14: Problem 4

Prove that for all integers \(n\) exactly one of \(n, 2 n-1\), and \(2 n+1\) is divisible by 3 .

Short Answer

Expert verified
For any given integer \(n\), exactly one of \(n\), \(2n - 1\), and \(2n + 1\) is divisible by 3. We have proven this by considering all possible forms of integer \(n\) (i.e. divisible by 3, one more than a multiple of 3, or two more than a multiple of 3) and showing that in each case, exactly one of the three given quantities is divisible by 3.

Step by step solution

01

Description of possible scenarios

We know that for any integer \(n\), it has to be one of the following three forms: it is either divisible by 3 (which can be written as \(n = 3k)\), one more than a multiple of 3 (\(n = 3k+1)\), or two more than a multiple of 3 (\(n = 3k+2)\). This is because \(n \% 3\), the remainder of \(n\) divided by 3, must be 0, 1, or 2.
02

Analyzing the first scenario: n is divisible by 3

Suppose \(n = 3k\). Then \(2n - 1 = 6k - 1\) and \(2n + 1 = 6k + 1\). Neither of \(2n - 1\) nor \(2n + 1\) is divisible by 3, as the remainder when divided by 3 is 2 for both cases. So in this case, \(n\) is the number that is divisible by 3.
03

Analyzing the second scenario: n equals one more than a multiple of 3

Suppose \(n = 3k + 1\). Then \(2n - 1 = 6k + 1\), and \(2n + 1 = 6k + 3\), which simplifies to \(2m = 3l\) where \(l = 2k + 1\). In this case, \(2n + 1\) is the number divisible by 3. Neither \(n\) nor \(2n - 1\) is divisible by 3 as their remainders when dividing by 3 is 1 and 2 respectively.
04

Analyzing the third scenario: n equals two more than a multiple of 3

Suppose \(n = 3k + 2\). Then \(2n - 1 = 6k + 3\), which simplifies to \(2n - 1 = 3l\) where \(l = 2k + 1\). So in this case, \(2n - 1\) is divisible by 3. Neither \(n\) nor \(2n + 1\) is divisible by 3 as their remainders when dividing by 3 is 2 and 1 respectively.

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Most popular questions from this chapter

Let \((R,+, \cdot)\) be a ring, with \(a \in R\). Define \(0 a=z, 1 a=a\), and \((n+1) a=n a+a\), for any \(n \in \mathbf{Z}^{*}\). (Here we are multiplying elements of \(R\) by elements of \(\mathbf{Z}\), so we have yet another operation that is different from the multiplications in either of \(\mathbf{Z}\) or \(R\).) For \(n>0\), we define \((-n) a=n(-a)\), so, for example, \((-3) a=3(-a)=2(-a)+(-a)=\) \([(-a)+(-a)]+(-a)=[-(a+a)]+(-a)=-[(a+a)+a]=-[2 a+a]=-(3 a) .\) For any \(a, b \in R\), and any \(m, n \in \mathbf{Z}\), prove that a) \(m a+n a=(m+n) a\) b) \(m(n a)=(m n) a\) c) \(n(a+b)=n a+n b\) d) \(n(a b)=(n a) b=a(n b)\) e) \((m a)(n b)=(m n)(a b)=(n a)(m b)\)

Consider the ring \(\left(\mathbf{Z}^{3}, \oplus, \odot\right)\) where addition and multiplication are defined by \((a, b, c) \oplus(d, e, f)=(a+d, b+e, c+f)\) and \((a, b, c) \odot(d, e, f)=(a d, b e, c f)\). (Here, for example, \(a+d\) and \(a d\) are computed by using the standard binary operations of addition and multiplication in \(\mathbf{Z}\).) Let \(S\) be the subset of \(\mathbf{Z}^{3}\) where \(S=\\{(a, b, c) \mid a=b+c\\}\). Prove that \(S\) is not a subring of \(\left(\mathbf{Z}^{3}, \oplus, \odot\right)\).

If \(S=\left\\{\left[\begin{array}{ll}a & 0 \\ 0 & a\end{array}\right] \mid a \in \mathbf{R}\right\\}\), then \(S\) is a ring under matrix addition and multiplication. Prove that \(\mathbf{R}\) is isomorphic to \(S\).

a) Construct a table (as in Example 14.18) for the isomorphism \(f: \mathbf{Z}_{20} \rightarrow \mathbf{Z}_{4} \times \mathbf{Z}_{5}\). b) Use the table from part (a) to compute the following in \(\mathbf{Z}_{\mathrm{n}}\). i) \((17)(19)+(12)(14)\) ii) \((18)(11)-(9)(15)\)

For the set \(R\) in Example \(14.4\), keep \(A \cdot B=A \cap B\), but define \(A+B=A \cup B\). Is \((R, \cup, \cap)\) a ring?
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