hello please help me answer these short questions.

1.

Click and drag the given steps (in the right) to their corresponding step names (in the left) to show that if a l b and bla, where a and b are integers, then a = b or a = -b. Step 1 Hence, a = a + c+ d. If a l b and bla, there are integers c and d such that b = ac and a = bd. Step 2 Thus, either c = d= 1 or c = d = -1. Thus, a = bor a = -b Step 3 If a l b and bla, there are integers c and d such that b = a + c and a = b + d. Because a # 0, it follows that cd = 1. Step 4 Hence, a = acd. ResetClick and drag the given steps (in the right) to their corresponding step names (in the left) to prove that if a, b, c, and dare integers, where a # 0 and b # 0, such that al c and bl d, then ab | cd. If ad = (cs)(bt) = cb(st). It follows that ab | cd. al c and bl d, there are integers s and t such that c = as and d = bt. Therefore, cd = (as)(bt) = ab(st). It follows that ab l cd. al c and bl d, there are integers s and t such that a = cs and b = dt.Click and drag the given steps (on the right) to their corresponding step names (on the left) to prove that if a | b and bl C, then al c. 5'9P 1 Suppose a | b and bi 0. By definition of divisibility, a | b means that b = atfor some integer t, and b | 0 means that c = bs for some integer s. We substitute the equation b = at into c = bs and get 0 = ats. Step 2 . . . . . _ . . . . ' By definition of diVISIbility, a = C(st), With ts being an integer, implies at c. Suppose a | b and bi c. By definition of divisibility, a | b means that a = btfor some integer I, and b | 0 means that b = as for some integer 5. Step 3 By definition of divisibility, c = a(ts), with ts being an integer, implies alc. We substitute the equation b = as into a = D! and get a = 051'. Reset Click and drag the given steps to their corresponding step names to prove that if a, b, and care integers, where a # 0 and c # 0, such that acl bc, then al b. Step 1 If acl bc, then there is an integer k such that bc = ack. Hence, K = c(b - a) - alb. Step 2 If ac l bc, then there is an integer k such that k = bc- ac. Hence, b = ak - al b.Identify the positive integers that are not relatively prime to 24' (Check all that apply') Check All That Apply DECIDE] Click and drag the given steps (on the right) to their corresponding step names (on the left) to show that a mod m = b mod m if a = b (mod m) where m be a positive integer. We know that b = qm + r for some nonnegetive r less than m. This means that m l a + b, say a + b = mc, so that a = b + mc. By definition, this means that b must also equal a mod m. we can write a = qm + r+ mc = (q + c)m+r. Assume that a = b (mod m). This means that m l a - b, say a - b = mc, so that a = b + mc.we can write a = qm + r + mc = (q+ c)m + r. Assume that a = b (mod m). This means that ml a - b, say a - b = mc, so that a = b + mc. By definition, this means that r must also equal a mod m. Let us compute a mod m. ResetClick and drag the given steps to their corresponding step names to prove that if n is an odd positive integer, then n- = 1 (mod 8). Step 1 Since 12 - 1 = k(k - 2) is a multiple of 8, n2 = 8/ + 1, where / is an integer. Thus, /2 = 1 (mod 8) by definition. Then, 12 = (k - 1)2 = k2 - 2k + 1 = k(k-2) + 1. Step 2 Since n2 - 1 = 4k(k + 1) is a multiple of 8, n2 = 8/+ 1, where / is an integer. Thus, n2 = 1 (mod 8) by definition. Step 3 Then, n2 = (2k + 1)2 = 42 + 4k+ 1 = 4k(k + 1) + 1. By definition of odd number, n = k - 1 for some integer k. Step 4 By definition of odd number, n = 2k + 1 for some integer k. Since either k or k - 2 is even, k(k - 2) is even. Thus, K(k- 2) is a multiple of 8. Since either k or k + 1 is even, k(k + 1) is even. Thus, 4k(k + 1) is a multiple of 8. ResetLet m be a positive integer. Show that a a b (mod m) if a mod m = b mod m. Drag the necessary statements and drop them into the appropriate blank to build your proof. Proof method: Proof's assumption(s): Implication(s) and deduction(s) resulting from the assumption(s): Conclusion(s) from implications and deductions: Example a - b= qm - q,m my (a - b) m | (a- b) Counter example a = qm+r a = q,m + rand b = q,m + r a mod m = b mod m for some integers q,, q, and an integer rwith 0 sr 1 and b >1. We must prove that 26b 1 is not prime. C] We will prove by contrapositive. Suppose n is not prime. Then, n = ab, for some integers 3 1, the factor 2'3 1 is greater than 1. E] Clearly, (2a(b 1) + 28(b- 2) + + 23+ 1) is less than 1' Since a