H Home Tools34★LO56Acrobat Reader File Edit View Sign Window HelpO)Additional Practic... xDiscrete Mathema...Additional Practice Problems on Proof.pdf11. If m is an even integer and n is an odd integer, then mn-m is divisible by 4.OCT10/ 12. Suppose a is a natural number > 1. Prove that if a|(b+15) and a|(b+22), then_a=7.2803. Suppose m is a natural number. Prove that if m² – 9 is prime, then_m= 4.4. For all positive real numbers a and b, if a< b then a ‹ b³ . You may assume that the product of twopositive numbers is positive, and the product of two negative numbers is also positive.479D5. We know that if gcd(a, b) = 1, then there exist integer solutions to the equation ax+by= 1. Prove thefollowing more general statement: Let gcd(a, b) = d. If dc, then there exist integer solutions to theequation_ax+ by= c. (Hint: as a starting point, consider the integers a/d and b/d. In particular, what istheir gcd?)StvLAO2fPdelodAddMon Oct 10 3:47 PMSearch 'Crop Page'Export PDFAdobe Export PDFConvert PDF Files to Wordor Excel OnlineSelect PDF FileAdditional ...n Proof.pdfConvert toOMicrosoft Word (*.docx)Document Language:English (U.S.) ChangeConvertEdit PDFCreate PDFSign InCommentCombine FilesW^XV

●●● H Home Tools 3 1999 Additional Practic... x Discrete Mathema... Additional Practice Problems on Proof.pdf 1 /1 @ Eu 1. If m is an even integer and n is an odd integer, then mn-m is divisible by 4. 2. Suppose a is a natural number> 1. Prove that if a (b+15) and a (b+ 22), then a = 7. 3. Suppose m is a natural number. Prove that if m² -9 is prime, then m = 4. OCT 10 4. For all positive real numbers a and b, if a< b then a

●●● H Home Tools 3 1999 Additional Practic... x Discrete Mathema... Additional Practice Problems on Proof.pdf 1 /1 @ Eu 1. If m is an even integer and n is an odd integer, then mn-m is divisible by 4. 2. Suppose a is a natural number> 1. Prove that if a (b+15) and a (b+ 22), then a = 7. 3. Suppose m is a natural number. Prove that if m² -9 is prime, then m = 4. OCT 10 4. For all positive real numbers a and b, if a< b then a

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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H Home Tools
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Acrobat Reader File Edit View Sign Window Help
O)
Additional Practic... x
Discrete Mathema...
Additional Practice Problems on Proof.pdf
1
1. If m is an even integer and n is an odd integer, then mn-m is divisible by 4.
OCT
10
/ 1
2. Suppose a is a natural number > 1. Prove that if a|(b+15) and a|(b+22), then_a=7.
280
3. Suppose m is a natural number. Prove that if m² – 9 is prime, then_m= 4.
4. For all positive real numbers a and b, if a< b then a ‹ b³ . You may assume that the product of two
positive numbers is positive, and the product of two negative numbers is also positive.
479
D
5. We know that if gcd(a, b) = 1, then there exist integer solutions to the equation ax+by= 1. Prove the
following more general statement: Let gcd(a, b) = d. If dc, then there exist integer solutions to the
equation_ax+ by= c. (Hint: as a starting point, consider the integers a/d and b/d. In particular, what is
their gcd?)
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Step 1

As per the guidelines I am supposed to do the first question among multiple questions. So here I am doing the first question only. To get the solution of the others please repost those separately as another question. Thank you.

Given m is an even integer.

So, m=2k, for any integer k

Given n is an odd integer.

So, n=2l+1, for any integer l

So, mn=4lk+2k

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Solved in 2 steps

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17. Evaluation of proofs See the instructions for Exercise (19) on page 100 from Section 3.1. (a) For each natural number n with n > 2, 2" > 1 +n. Proof. We let k be a natural number and assume that 2k > 1 + k. Multiplying both sides of this inequality by 2, we see that 2k+1 > 2+ 2k. However, 2 + 2k > 2 + k and, hence, 2k+1 > 1+ (k + 1). By mathematical induction, we conclude that 2" >1+ n.
24. Binomial Multiplication Patterns in Pascal's Triangle a.) What is (x+1)°? (Hint: anything to the 0 power is b.) What is (x+1)'? c.) What is (x+1)²? Caution: it is not x2+12, but is gotten by multiplying (x + 1)(x + 1). d.) Where can the coefficients and constants of each answer be found in Pascal's triangle? The coefficient is the number in front of x, and the constant is the added on number. For x + 1, the coefficient in front of the x is a 1, so we have 1x + 1. Where do you see 1 1 in the triangle? Where do you see the coefficients and constants for your answer to part c? 4 1 10 10 1 6 I 15 I 20 I 15 7 I 21 I 35 56 I 70 I 56 I 28 36 I 84 1126 | 126 I 84 I 36 1 6 1 1 35 I 21 I 7 1 8 28 8 1 9 e.) What is (x+1)³? Hint: multiply the result of part c by (x+1), then combine like terms. Explain how you know you are correct using Pascal's triangle. f.) Make a conjecture as to what (x+1)4 will equal, using Pascal's triangle. You do not have to multiply it all out to check, but instead,…
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DO NOT COPY FROM OTHER SITES. THANK YOU. PLEASE PROVE IT.
One (1.5-1.8, 2.1, 2. X Ô https://courses.campbellsville.edu/mod/quiz/attempt.php?attempt=2249428&cmid= Resources Solve the inequality for x, assuming that p, b. and d are positive constants. p(bx - d) 2 bd O a. p(d + b) db Ob. b(p + d) pd d(b - p) pb O d. d(p + b) pb p(p - b) Oe. db