Exercise 4.1.4 (Integers have no zero divisors) Let \(a\) and \(b\) be integers such that \(ab = 0\). Then either \(a = 0\) or \(b = 0\) (or both)
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\(\Vdash\) \(a \neq 0_Z\) and \(b \neq 0_Z\)
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\(\equiv\) { \(a = x — y\), \(b = z — w\), \(0_Z = k — k\) }
- \(x — y \neq k — k\) and \(z — w \neq k — k\)
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\(\equiv\) { Definition 4.1.1 }
- \(x + k \neq y + k\) and \(z + k \neq w + k\)
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\(\equiv\) { Proposition 2.2.6 }
- \(x \neq y\) and \(z \neq w\)
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\(\equiv\) { Without loss of Generality, \(x > y\) and \(z > w\) and \(c \neq 0_N\) and \(d \neq 0_N\) }
- \(x = y + c\) and \(z = w + d\)
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\(\Vdash\) \(ab\)
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\(=\) { \(a = x — y\), \(b = z — w\) }
- \((xz + yw) — (xw + yz)\)
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\(=\) { \(x = y + c\) and \(z = w + d\) }
- \((yw + yd + cw + cd + yw) — (yw + cw + yw + yd)\)
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\(=\) { Definition 4.1.1 }
- \(cd — 0_N\)
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\(\neq\) { By contrapositive of Lemma 2.3.3, \(c \neq 0_N\) and \(d \neq 0_N\), Definition 4.1.1 }
- \(0_Z\)
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\(\square\)