Let's solve each problem step by step with explanations. --- # Problem 5 **Prove that if \( i \) is an integer, then \( 6i - 7 \) is an odd integer.** ## Step-by-step Solution ### 1. **Let \( i \) be any integer** Let \( i \) be any integer (\( i \in \mathbb{Z} \)). ### 2. **Express \( 6i \)** \( 6i \) is always even because any integer multiplied by 6 (an even number) is even. Let \( 6i = 2k \) for some integer \( k \). ### 3. **Subtract 7** \( 6i - 7 = (2k) - 7 = 2k - 7 \) ### 4. **Check the parity (odd/even)** Recall: Odd integer = \( 2m + 1 \) for some integer \( m \). Let’s write \( 2k - 7 \) as \( 2m + 1 \): \( 2k - 7 = 2k - 8 + 1 = 2(k - 4) + 1 \) Since \( k \) is integer, \( (k-4) \) is also integer, say \( m = k-4 \). ### 5. **Conclusion** Thus, \( 6i - 7 \) can be written as \( 2m + 1 \) for some integer \( m \), so it is **odd**. **Final Answer:** If \( i \) is an integer, then \( 6i - 7 \) is an odd integer. --- # Problem 6 **Prove that if \( a, b, c \) are odd integers, then \( a + b + c \) is an odd integer.** ## Step-by-step Solution ### 1. **Express \( a, b, c \) as odd integers** Let \( a = 2m + 1 \), \( b = 2n + 1 \), \( c = 2p + 1 \), for integers \( m, n, p \). ### 2. **Sum \( a + b + c \)** \( a + b + c = (2m + 1) + (2n + 1) + (2p + 1) \) \( = 2m + 2n + 2p + 3 \) \( = 2(m + n + p) + 3 \) ### 3. **Express as odd** \( 2(m + n + p) + 3 = 2(m + n + p + 1) + 1 \) Let \( q = m + n + p + 1 \), which is an integer. So, \( a + b + c = 2q + 1 \), which is odd. **Final Answer:** If \( a, b, c \) are odd integers, then \( a + b + c \) is an odd integer. --- # Problem 7 ### a. **If \( 2x \) is rational, then \( x - 3 \) is rational** #### Solution: 1. If \( 2x \) is rational, \( 2x = r \) for some rational \( r \). 2. \( x = \frac{r}{2} \), which is rational (since rationals are closed under division by nonzero integers). 3. \( x - 3 = \frac{r}{2} - 3 = \frac{r - 6}{2} \), which is rational (since rationals are closed under addition/subtraction). **Conclusion:** If \( 2x \) is rational, then \( x - 3 \) is rational. --- ### b. **If \( x - 3 \) is rational, then \( x/5 \) is rational** #### Solution: 1. If \( x - 3 \) is rational, \( x - 3 = r \) for some rational \( r \). 2. \( x = r + 3 \), which is rational (sum of rationals). 3. \( x/5 = (r + 3)/5 = r/5 + 3/5 \), which is rational (rationals closed under addition/division by nonzero integers). **Conclusion:** If \( x - 3 \) is rational, then \( x/5 \) is rational. --- ### c. **If \( x/5 \) is rational, then \( 2x \) is rational** #### Solution: 1. If \( x/5 \) is rational, \( x/5 = r \) for some rational \( r \). 2. \( x = 5r \), which is rational (product of rational and integer). 3. \( 2x = 2(5r) = 10r \), which is rational. **Conclusion:** If \( x/5 \) is rational, then \( 2x \) is rational. --- ## **Summary Table** | Problem | Final Answer | |---------|--------------| | 5 | If \( i \) is integer, \( 6i-7 \) is odd | | 6 | If \( a, b, c \) are odd, \( a+b+c \) is odd | | 7a | If \( 2x \) is rational, \( x-3 \) is rational | | 7b | If \( x-3 \) is rational, \( x/5 \) is rational | | 7c | If \( x/5 \) is rational, \( 2x \) is rational |