Prove the following properties of \(\mathbb{Z}\) and \(\mathbb{Z}_{+}\) :
\(a, b \in \mathbb{Z}_{+} \Rightarrow a + b \in \mathbb{Z}_{+}\) . [Hint: Show that given \(a \in \mathbb{Z}_{+}\), the set \(X = \{x \mid x \in \mathbb{R} \text{ and } a + x \in \mathbb{Z}_{+}\}\) is inductive.]
\(a, b \in \mathbb{Z}_+ \Rightarrow a \cdot b \in \mathbb{Z}_+\) .
Show that \(a \in \mathbb{Z}_+ \Rightarrow a - 1 \in \mathbb{Z}_+ \cup \{0\}\) . [Hint: Let \(X = \{x \mid x \in \mathbb{R} \text{ and } x - 1 \in \mathbb{Z}_+ \cup \{0\}\) ; show that \(X\) is inductive.]
\(c, d \in \mathbb{Z} \Rightarrow c + d \in \mathbb{Z}\) and \(c - d \in \mathbb{Z}\) . [Hint. Prove it first for \(d = 1\) .]
\(c, d \in \mathbb{Z} \Rightarrow c \cdot d \in \mathbb{Z}\) .
Solution
Fix \(a\in\mathbb Z_+\), and define \[ X=\{x\in\mathbb R:a+x\in\mathbb Z_+\}. \] Since \(a+1\in\mathbb Z_+\), we have \(1\in X\). If \(x\in X\), then \(a+x\in\mathbb Z_+\), so \[ a+(x+1)=(a+x)+1\in\mathbb Z_+, \] hence \(x+1\in X\). Thus \(X\) is inductive, so \(\mathbb Z_+\subset X\). In particular, for every \(b\in\mathbb Z_+\), \[ a+b\in\mathbb Z_+. \]
Fix \(a\in\mathbb Z_+\), and let \[ Y=\{x\in\mathbb R:ax\in\mathbb Z_+\}. \] Since \(a\cdot1=a\in\mathbb Z_+\), \(1\in Y\). If \(x\in Y\), then by part (a) \[ a(x+1)=ax+a\in\mathbb Z_+, \] so \(x+1\in Y\). Thus \(Y\) is inductive, whence \(b\in Y\) for every \(b\in\mathbb Z_+\). Therefore \(ab\in\mathbb Z_+\).
Let \[ X=\{x\in\mathbb R:x-1\in\mathbb Z_+\cup\{0\}\}. \] Since \(1-1=0\), \(1\in X\). If \(x\in X\), then either \(x-1=0\), in which case \(x=1\in\mathbb Z_+\), or \(x-1\in\mathbb Z_+\), in which case \[ x=(x-1)+1\in\mathbb Z_+. \] Thus in either case \[ (x+1)-1=x\in\mathbb Z_+, \] so \(x+1\in X\). Hence \(X\) is inductive and \(\mathbb Z_+\subset X\), proving \[ a-1\in\mathbb Z_+\cup\{0\} \qquad(a\in\mathbb Z_+). \]
First note that adding or subtracting \(1\) preserves \(\mathbb Z\). For positive integers this follows from inductivity and part (c); for \(0\) it is immediate; for a negative integer \(-a\), \[ -a+1=-(a-1),\qquad -a-1=-(a+1), \] and parts (a),(c) show these are integers. Every \(d\in\mathbb Z\) is \(0\), a positive integer, or the negative of one. Repeatedly adding or subtracting \(1\), and using induction on the positive integer \(|d|\), therefore gives \[ c+d\in\mathbb Z, \qquad c-d\in\mathbb Z. \]
If either factor is zero, the product is zero. Otherwise write \[ c=\varepsilon a,\qquad d=\delta b, \] with \(a,b\in\mathbb Z_+\) and \(\varepsilon,\delta\in\{1,-1\}\). Part (b) gives \(ab\in\mathbb Z_+\), and the sign laws of E-MUN-4-1(e) give \[ cd=(\varepsilon\delta)ab\in\mathbb Z. \]