0E: Exercises
- Page ID
- 131044
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Define \(h : \mathbb{Z} \rightarrow \mathbb{Z} \) by \(h(x) = x^2+4 \). Determine (with reasons) whether or not \(h \) is one−to−one and whether or not \(h \) is onto.
Suppose \(f : A \rightarrow B \) and \(g : B \rightarrow C \) are functions. Prove or disprove the following statements:
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If \(g \circ f \) is one-to-one then \(g \) is one-to-one.
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If \(g \circ f \) is one-to-one and \(f \) is onto, then \(g \) is one-to-one.
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If \(g \circ f \) is onto and \(g \) is one-to-one, then \(f \) is onto.
Determine whether or not each of the following binary relations \(R \) on the given set \(A \) is reflexive, symmetric, antisymmetric, or transitive. If a relation has a certain property, prove this is so; otherwise, provide a counterexample to show that it does not. If \(R \) is an equivalence relation, describe the equivalence classes of \(A \).
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Let \(S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \). Define a relation \(R \) on \(A = S \times S \) by \((a, b) R (c, d) \) if and only if \(10a + b \le 10c + d \).
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Let \(A = \mathbb{Z} \backslash \{0\} \). Define a relation \(R \) on \(A \), by \(a R b \) if and only if \( ab > 0 \).
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Define a relation \(R \) on \(A = \mathbb{Z} \) by \(a R b \) if and only if \(4 | (3a + b) \).
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Define a relation \(R \) on \(A = \mathbb{Z} \) by \(a R b \) if and only if \(3 | (a^2 - b^2 ) \).
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Let \(A = \mathbb{R} \), If \(a,b \in \mathbb{R} \), define \(a R b \) if and only if \(a - b \in \mathbb{Z} \).
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Define a relation \(R \) on the set \(\mathbb{Z} \times \mathbb{Z} \) by \((a, b) R (c, d) \) if and only if \(ac = bd \).
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Define a relation \(R \) on \(\mathbb{Z} \) by \(a R b \) if and only if \(2 \mid a^2+b\).
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Let \(A = \mathbb{R} \), If \(a,b \in \mathbb{R} \), define \(a R b \) if and only if \(a - b \in \mathbb{Q} \).
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Let \(A=\mathbb{R} \times \mathbb{R} \), If \((x,y),(x_1,y_1) \in \mathbb{R}\times \mathbb{R}\), define \((x,y) \, R \, (x_1,y_1)\) if and only if \( x^2+y^2=x_1^2+y_1^2.\) -
Define a relation \(R \) on \(\mathbb{Z} \) by \(a R b \) if and only if \(5 | (2a + 3b) \).
- Answer
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- \(R \) is reflexive on \(\mathbb{Z} \).
Proof:
Let \(a \in \mathbb{Z} \).
We shall show that \(a R a \), specifically \(5|(2a+3a) \).
Consider that \(2a+3a = 5a \) and \(a \in \mathbb{Z} \).
Thus \(5|(2a+3a) \), and \(aRa \).
Therefore \(R \) is reflexive on \(\mathbb{Z} \).◻
2. \(R \) is symmetric on \(\mathbb{Z} \).
Proof:
Let \(a,b \in \mathbb{Z} \) s.t. \(5|(2a+3b \).
Thus \(2a+3b=5m \) for some \(m\in \mathbb{Z} \).
We will show that \(3a+2b=5k \) for some \(k \in \mathbb{Z} \).
Consider that \(-2a-3b=-5m \) for some \(m\in \mathbb{Z} \).
Then \(5(a+b) -2a-3b=5(a+b)-5m \).
Thus \(3a+2b=5(a+b-m) \) where \(a+b-m=k \in \mathbb{Z} \).
Hence \(5|(3a+2b) \) and \(bRa \).
Since \(bRa \), \(R \) is symmetric on \(\mathbb{Z} \).
3. \(R \) is not antisymmetric on \(\mathbb{Z} \).
Counter Example:
Let \(a=0 \) and \(b=5 \).
Then \(5|(2(0)+3(5)) \) and \(5|(2(5)+3(0) \).
However, since \(0 \ne 5 \) \(R \) is not antisymmetric on \(\mathbb{Z} \).◻
4. \(R \) is transitive on \(\mathbb{Z} \).
Let \(a,b,c \in \mathbb{Z} \) s.t. \(aRb \), \(5|(2a+3b) \) and \(bRa \), \(5|(2b+3c) \).
We will show that \(aRc \), \(5|(2a+3c) \).
Since \(5|(2a+3b) \), \(2a+3b=5(k) \) for some \(k \in \mathbb{Z} \).
Since \(5|(2b+3c) \), \(2b+3c=5(m) \) for some \(m \in \mathbb{Z} \).
Consider \(2a+3c=(2a+3b)+(2b+3c)-5b \)
\(=5(k)+5(m)-5(b) \)
\(=5(k+m-b) \), where \(k+m-b \in \mathbb{Z} \).
Hence \(5|2a+3c \) and \(aRc \).
Hence \(R \) is transitive on \(\mathbb{Z} \).◻
Since \(R \) is reflexive, symmetric and transitive on \(\mathbb{Z} \), \(R \) is an equivalence relation.
Te equivalence classes of \(aRb \) iff \(5 | (2a + 3b) \) are \([0], [1], [2], [3] \) and \([4] \).
Let \(a \in \mathbb{Z} \), then \([a]=\{a\in \mathbb{Z}:x \sim a\} \).
\([0]=\{x \in \mathbb{Z}: x\sim 0\} \)
\(=\{x \in \mathbb{Z}: 5|(2x+3(0)) \} \)
\(=\{x \in \mathbb{Z}: 5|2x \} \)
\(=\{x \in \mathbb{Z}: 2x=5m, m\in \mathbb{Z} \} \)
\(=\{\ldots, -10,-5,0,5,10,\ldots\} \).
\([1]=\{x \in \mathbb{Z}: x\sim 1\} \)
\(=\{x \in \mathbb{Z}: 5|(2x+3(1)) \} \)
\(=\{x \in \mathbb{Z}: 5|(2x+3) \} \)
\(=\{x \in \mathbb{Z}: 2x+3=5m, m\in \mathbb{Z} \} \)
\(=\{x \in \mathbb{Z}: 2x=5m-3, m\in \mathbb{Z} \} \)
\(=\{\ldots, -9,-4,1,6,11,\ldots\} \).
\([2]=\{x \in \mathbb{Z}: x\sim 2\} \)
\(=\{x \in \mathbb{Z}: 5|(2x+3(2)) \} \)
\(=\{x \in \mathbb{Z}: 5|(2x+6) \} \)
\(=\{x \in \mathbb{Z}: 2x+6=5m, m\in \mathbb{Z} \} \)
\(=\{x \in \mathbb{Z}: 2x=5m-6, m\in \mathbb{Z} \} \)
\(=\{\ldots, -8,-3,2,7,12,\ldots\} \).
\([3]=\{x \in \mathbb{Z}: x\sim 3\} \)
\(=\{x \in \mathbb{Z}: 5|(2x+3(3)) \} \)
\(=\{x \in \mathbb{Z}: 5|(2x+9) \} \)
\(=\{x \in \mathbb{Z}: 2x+9=5m, m\in \mathbb{Z} \} \)
\(=\{x \in \mathbb{Z}: 2x=5m-9, m\in \mathbb{Z} \} \)
\(=\{\ldots, -7,-2,3,8,13,\ldots\} \).
\([4]=\{x \in \mathbb{Z}: x\sim 4\} \)
\(=\{x \in \mathbb{Z}: 5|(2x+3(4)) \} \)
\(=\{x \in \mathbb{Z}: 5|(2x+12) \} \)
\(=\{x \in \mathbb{Z}: 2x+12=5m, m\in \mathbb{Z} \} \)
\(=\{x \in \mathbb{Z}: 2x=5m-12, m\in \mathbb{Z} \} \)
\(=\{\ldots, -6,-1,4,9,14,\ldots\} \)