Review for CS1050

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Review for CS1050
2
Review Questions

• Without using truth tables, prove that ?(p?q) ?
  ?q is a tautology.
• Prove that the sum of an even integer and an odd
  integer is always odd.
• Use mathematical induction to prove that
  
  
• 
  whenever n?Z.

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Review Questions

• Prove that fn1fn-1 fn2 (-1)n whenever n is a
  positive integer.
• Without using set membership tables, prove or
  disprove that if A,B and C are sets, then A
  (B?C) (A-B) ? (A-C).
• Using the definition of Big Oh prove that n! is
  O(nn).
• Prove that (3,5,7) is the only prime triple,
  i.e., the only set of three consecutive odd
  integers gt 1 that are all prime.

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Review Questions

• Suppose f Z ? Z where f(n) 4n 1 and Z is
  the set of all non-negative integers.
• Is f one-to-one?
• Is f onto?
• Suppose f Z ? Z where f(n) 4n2 1 and Z is
  the set of all non-negative integers
• Is f one-to-one?
• Is f onto?
• Consider the relation R (x,y) x and y are
  both negative or both nonnegative integers
• Is R reflexive?
• Is R symmetric?  
• Is R transitive?

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Without using truth tables, prove that ?(p?q) ?
?q is a tautology.

• ?(p?q) ? ?q ?
• ?(?p?q) ? ?q ? Misc. T6 (Implication
  equivalence)
• ??(?p?q) ? ?q ? Misc. T6 (Implication
  equivalence)
• (?p?q) ? ?q ? Double Negation
• ?p? (q ? ?q) ? Associative
• ?p? T ? Misc. T6 (Or tautology)
• T ? Domination
•  

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Prove that the sum of an even integer and an odd
integer is always odd.

• Let e be an even integer and f be an odd integer.
  Then there exists i,j ? Z such at e 2i and
  f2j1.
•  (Why not e 2j and f2j1?)
• Then ef 2i2j1 2(ij) 1 which is odd
  since ij must be an integer.

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Use mathematical induction to prove that


whenever n?Z.

• Basic Case Let n 1, then
•  
• Inductive Case
• Assume that the expression is true for n, i.e.,
  that
• then we must show that

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Prove that fn1fn-1 fn2 (-1)n whenever n is a
positive integer.

• Basis Step For n1 we have f2f0 f12 10
  12 -1 which is (-1)1
•  
• Inductive Step
• Assume the inductive hypothesis that fn1fn-1
  fn2 (-1)n . We must show that
• that fn11fn1-1 fn12 (-1)n1

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Then for n1 we have that fn11fn1-1 fn12
fn2fn fn12 (fn1fn )fn- fn12 (fn1fn)
fn2- fn12 (fn1)(fn - fn1) fn2
(-fn1)(fn1-fn) fn2 But we know that fn1-fn
fn-1 so (-fn1)(fn-1) fn2 -(fn1fn-1- fn2
) -(-1)n by the inductive hypothesis which is
(-1)n1
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Without using set membership tables, prove or
disprove that if A,B and C are sets, then A
(B?C) (A-B) ? (A-C).

• Proof
• We must show that A (B?C) ? (A-B) ? (A-C) and
  that (A-B) ? (A-C) ? A (B?C).
•   
• First we will show that A (B?C) ? (A-B) ?
  (A-C). Let e be an arbitrary element in A
  (B?C). Then e? A but e?(B?C). Since e?(B?C),
  then either e?B or e?C or both. If e?A but e?B,
  then e?(A-B). If e?A but e?C, then e?(A-C).
  Since e is either an element of (A-B) or (A-C),
  then e must be in their union.. Therefore e ?
  (A-B) ? (A-C).
•  

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Now we will show that (A-B) ? (A-C) ? A
(B?C).   Let e be an arbitrary element in (A-B) ?
(A-C). Then either e is in (A-B) or e is in
(A-C) or e is in both. If e is in (A-B), then e
? A but e?B. If e? B then e?(B?C). Therefore
e?A-(B?C). If e is in (A-C), then e ? A but e?C.
If e? C then e?(B?C). Therefore e?A-(B?C).  
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Using the definition of Big Oh prove that n! is
O(nn).

• Proof We must show that ? constants C?N and k?R
  such that n! ? Cnn whenever n gt k.
•  
• n! n(n-1)(n-2)(n-3)(3)(2)(1)
• ? n(n)(n)(n)(n)(n)(n) n times
• nn
•  
• So choose k 0 and C 1
•  

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Prove that (3,5,7) is the only prime triple,
i.e., the only set of three consecutive odd
integers gt 1 that are all prime

• We must show that, for any three consecutive odd
  integers, at least one is divisible by some other
  number. We will show that for every three
  consecutive odd integers, one of them is
  divisible by 3.
• Proof
• Any three consecutive odd integers can be written
  as 2k1, 2k3, and 2k5 for k?Z.
• For any k?Z, there are three possible cases,
• k 3j, k 3j1, k 3j 2 for j ? Z.

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• Case 1 k 3j.
• Then three consecutive odd integers look like
• 2k1 2(3j) 1 3(2j) 1
• 2k3 2(3j) 3 3(2j) 3
• 3(2j1) which is divisible by 3 since 2j1 is
  an integer
• 2k5 2(3j) 5 3(2k) 5 3(2k1) 2

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• Case 2 k 3j1
• Then three consecutive odd integers look like
• 2k1 2(3j1) 1 3(2j1) which is divisible by
  3 since 2j1 is an integer.
• 2k3 2(3j1) 3 3(2j1) 2
• 2k5 2(3j1) 5 3(2j2) 1

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• Case 3 k 3j2
• Then three consecutive odd integers look like
• 2k1 2(3j2) 1 3(2j1) 2 2k3 2(3j2) 3
  3(2j2) 1
• 2k5 2(3j2) 5 3(2j3) which is divisible
  by 3 since 2j3 is an integer.
• The only number divisible by 3 and that is prime
  is 3 so 3,5,7 is the only possible prime
  triple.

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Suppose fZ ? Z where f(n) 4n 1 and Z is
the set of all non-negative integers.

• Is f one-to-one?
• Is f onto?
• Suppose f Z ? Z where f(n) 4n2 1 and Z is
  the set of all non-negative integers
•  
• Is f one-to-one?
• Is f onto?
•  

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Consider the relation R (x,y) x and y are
both negative or both nonnegative integers

• Is R reflexive?
• Yes. Since (x,x) is in R if x is positive or
  negative.
• Is R symmetric?  
• Yes. Since (x,y) in R means that both x and y
  are negative or nonnegative so the same is true
  of their reverse order (y,x).
• Is R transitive?
• Yes. Since (x,y) and (y,z) both in R implies
  that x,y and z must all have the same sign.