MATH Section 5.2
Prove the statement that n cents of postage can be formed with just 4-cent and 11-cent stamps using strong induction, where n ≥ 30. Let P(n) be the statement that we can form n cents of postage using just 4-cent and 11-cent stamps. To prove that P(n) is true for all n ≥ 30, identify the inductive hypothesis used in the strong induction. (You must provide an answer before moving to the next part.) Multiple Choice P(j) is true for all j with 30 ≤ j < k, where k is a fixed integer greater than or equal to 33. P(j) is true for all j with 30 ≤ j ≤ k, where k is a fixed integer greater than or equal to 33. Correct P(j) is true for all j with 30 ≤ j ≤ k, where k is a fixed integer greater than or equal to 30. P(j) is true for all j with 30 < j ≤ k, where k is a fixed integer greater than or equal to 33. P(j) is true for all j with 30 ≤ j ≤ k, where k is a fixed integer greater than or equal to 34. Explanation For the inductive step, assume the inductive hypothesis, that P(j) is true for all j with 30 ≤ j ≤ k, where k is an arbitrary integer greater than or equal to 33. We want to show that P(k + 1) is true.
P(j) is true for all j with 30 ≤ j ≤ k, where k is a fixed integer greater than or equal to 33.
Find the flaw with the following "proof" that every postage of 3 cents or more can be formed using just 3-cent and 4-cent stamps. Basis Step: We can form postage of 3 cents with a single 3-cent stamp, and we can form postage of 4 cents using a single 4-cent stamp. Inductive Step: Assume that we can form postage of j cents for all nonnegative integers j with j ≤ k using just 3-cent and 4-cent stamps. We can then form postage of k + 1 cents by replacing one 3-cent stamp with a 4-cent stamp or by replacing two 4-cent stamps by three 3-cent stamps. Which of these statements describes a flaw in the proof? Multiple Choice The inductive step fails for k = 6 because no 4-cent stamps are used to form 6 cents of postage, so neither proposed replacement is possible. The inductive step fails for k = 8 because no 3-cent stamps are used to form 8 cents of postage, so neither proposed replacement is possible. The inductive step fails for k = 3 because 3 cents of postage includes neither a 4-cent stamp nor two 4-cent stamps; so, neither proposed replacement is possible. The inductive step fails for k = 4 because 5 cents of postage includes neither a 3-cent stamp nor two 4-cent stamps; so, neither proposed replacement is possible. Explanation The statement is "every postage of 3 cents or more can be formed using just 3-cent and 4-cent stamps." Firstly, the statement itself is not true. Further, the inductive step is invalid for k = 4. We cannot increase the postage from 4 cents to 5 cents using either of the replacements indicated, because there is no 3-cent stamp present and there is only one 4-cent stamp present. The flaw is in the inductive step. The inductive hypothesis only guarantees that postage for k stamps can be formed from 3-and 4-cent stamps, but not that there will be more than one 4 cent stamp. The argument in the inductive step requires that if there are no 3-cent stamps, then there are at least two 4-cent stamps, which is not the case. This mistake was encouraged by the additional, formal mistake of omitting the quantification "for some arbitrary k ≥ 4" from the inductive hypothesis.
The inductive step fails for k = 4 because 5 cents of postage includes neither a 3-cent stamp nor two 4-cent stamps; so, neither proposed replacement is possible.
Prove by strong induction that all amounts of postage greater than or equal to 18 cents can be formed using just 3-cent and 10-cent stamps. You must show all steps.
basis step: P(18) = 3 + 3 + 3 + 3 + 3 + 3 P(19) = 10 + 3 + 3 + 3 P(20) = 10 + 10 inductive hypothesis:using just 3-cent and 10-cent stamps we can form j cents postage for all j with 18 being less than or equal to j, and j being less than or equal to k (18 <= j <= k), where k is greater than or equal to 20 (k >= 20). inductive step:in the inductive step show, assuming the inductive hypothesis, that we can form k+1 cents postage using just 3-cent and 10-cent stamps. work for inductive step:if k-2 is greater than or equal to 18 (k-2 >= 18), then we can form a postage of k-2 cents. This means that P(k-2) is true.if you add a 3-cent stamp to P(k-2)-cents then you would reach k+1 cents postage (-2+3=1). therefore P(k+1) is true. This completes the inductive step. The basis step is also complete, so by using strong induction for any integer n greater than or equal to 18 (n>=18), P(n) is true. basis step: P(18) = 3 + 3 + 3 + 3 + 3 + 3 P(19) = 10 + 3 + 3 + 3 P(20) = 10 + 10 inductive hypothesis:using just 3-cent and 10-cent stamps we can form j cents postage for all j with 18 being less than or equal to j, and j being less than or equal to k (18 <= j <= k), where k is greater than or equal to 20 (k >= 20). inductive step:in the inductive step show, assuming the inductive hypothesis, that we can form k+1 cents postage using just 3-cent and 10-cent stamps. work for inductive step:if k-2 is greater than or equal to 18 (k-2 >= 18), then we can form a postage of k-2 cents. This means that P(k-2) is true.if you add a 3-cent stamp to P(k-2)-cents then you would reach k+1 cents postage (-2+3=1). therefore P(k+1) is true. This completes the inductive step. The basis step is also complete, so by using strong induction for any integer n greater than or equal to 18 (n>=18), P(n) is true.
Prove the statement that n cents of postage can be formed with just 4-cent and 11-cent stamps using strong induction, where n ≥ 30. Let P(n) be the statement that we can form n cents of postage using just 4-cent and 11-cent stamps. To prove that P(n) is true for all n ≥ 30, identify the proper basis step used in strong induction. (You must provide an answer before moving to the next part.) Multiple Choice 30 = 11 + 11 + 4 + 4, 31 = 11 + 5(4), 32 = 8(4) P(n) is true for n = 30, 31, 32. 31 = 11 + 5(4), 32 = 8(4), 33 = 3(11), 34 = 3(4) + 11 + 11 P(n) is true for n = 31, 32, 33, 34. 31 = 11 + 5(4), 32 = 8(4), 33 = 3(11) P(n) is true for n = 31, 32, 33. 30 = 11 + 11 + 4 + 4, 31 = 11 + 5(4), 32 = 8(4), 33 = 3(11) P(n) is true for n = 30, 31, 32, 33. Correct 30 = 11 + 11 + 4 + 4, 31 = 11 + 5(4), 32 = 8(4), 33 = 3(11), 34 = 3(4) + 11 + 11 P(n) is true for n = 30, 31, 32, 33, 34. Explanation P(n) is the same as in part (b). To prove that P(n) is true for all n ≥ 30, we check for the basis step that 30 = 11 + 11 + 4 + 4, 31 = 11 + 5(4), 32 = 8(4), and 33 = 3(11).
30 = 11 + 11 + 4 + 4, 31 = 11 + 5(4), 32 = 8(4), 33 = 3(11) P(n) is true for n = 30, 31, 32, 33.
Suppose you have only 2-dollar and 5-dollar bills. Identify the amounts that can be formed by using just 2-dollar and 5-dollar bills. (You must provide an answer before moving to the next part.) All amounts except $1 and $3CorrectAll amounts starting from $3Only $2 and $5All amounts starting from $1 Explanation We can form the following amounts of money as indicated:$2 = $2$4 = $2 + $2$5 = $5$6 = $2 + $2 + $2$7 = $5 + $2, and so on. We cannot form $1 and $3 using just $2 and $5 bills.
All amounts except $1 and $3
Prove the statement that n cents of postage can be formed with just 4-cent and 11-cent stamps using strong induction, where n ≥ 30. How does the inductive hypothesis used in strong induction differ from that in the inductive hypothesis for a proof using mathematical induction? Multiple Choice In this proof, the inductive hypothesis was that P(j) was true for all values of j between 30 and k inclusive, rather than just that P(30) was true. In this proof, the inductive hypothesis was that P(j) was true for all values of j between 30 and k , rather than just that P(30) was true. In this proof, the inductive hypothesis was that P(j) was true for all values of j between 30 and k inclusive, rather than just that P(35) was true. In this proof, the inductive hypothesis was that P(j) was true for all values of j between 32 and k inclusive, rather than just that P(32) was true. In this proof, the inductive hypothesis was that P(30) was true, rather than just P(j) was true for all values of j between 30 and k inclusive Explanation In the inductive hypothesis, P(j) was true for just P(30). In the strong induction, P(j) was true for all values of j between 30 and k inclusive. This is the difference between the inductive hypothesis and the strong induction in this proof.
In this proof, the inductive hypothesis was that P(j) was true for all values of j between 30 and k inclusive, rather than just that P(30) was true.
Suppose you have only 2-dollar and 5-dollar bills. Let P(n) be the statement "We can form n dollars using just 2-dollar and 5-dollar bills." Click and drag the steps in the correct order to prove using strong induction that P(n) is true for every positive integer n ≥ 5. Explanation The amount $5 can be formed using just one 5-dollar bill, and the amount $6 can be formed by using just three 2-dollar bills. Thus, the basis step is true for k = 5 and k = 6.Assume that P(j) is true for all j with 5 ≤ j ≤ k, where k is fixed integer greater than or equal to 6.We have to show that P(k + 1) is true.Because k - 1 ≥ 5, we know that P(k - 1) is true, that is, we can form k - 1 dollars using just 2-dollar and 5-dollar bills.Add another 2-dollar bill, and we can form k + 1 dollars, as desired.Therefore, by strong induction, P(n) is true for all positive integers n ≥ 5.
correct blocks and order (going down): basis step: The amount $5 can be formed using just one 5-dollar bill, and the amount $6 can be formed by using just three 2-dollar bills. Thus, the statement holds true for k = 5 and k = 6. inductive step: Assume that P(j) is true for all j with 5 ≤ j ≤ k, where k is fixed integer greater than or equal to 6. Because k - 1 ≥ 5, we know that P(k - 1) is true, that is, we can form k - 1 dollars using just 2-dollar and 5-dollar bills Add another 2-dollar bill, and we have formed k + 1 dollars, as desired.
Prove the statement that n cents of postage can be formed with just 4-cent and 11-cent stamps using strong induction, where n ≥ 30. Click and drag the given steps (in the right) to the corresponding step names given (in the left) to show that if P(j) is true for all j ≤ k, then P(k + 1) must also be true. (You must provide an answer before moving to the next part.) Explanation Because k − 3 ≥ 30, we know that P(k − 3) is true, that is, that we can form k − 3 cents of postage. Put one more 4-cent stamp on the envelope, and we have formed k + 1 cents of postage, as desired.
correct blocks and order (going down): step 1: Because k - 3 ≥ 30, we know that P(K - 3) is true, that is, that we can form k - 3 cents of postage. step 2: Put one more 4-cent stamp on the envelope, and we have formed k + 1 cents of postage, as desired.
