Download Math 2534 Test 2 Solution Summer 2014

Math 2534 Test 2 Solution Summer 2014
Problem 1: (12pts)
Using a direct method prove the following using definitions only (state them correctly.)
Theorem: Let a be an integer. If 2 (a  1), then a is odd.
Proof: Given 2 (a  1) , by definition of divisible there exist an integer q so that 2q = a – 1.
Solving for a we have that a = 2q + 1 and is odd by definition of odd.
Problem 2: (18 pts) Using proof by contrapositive prove the following. (use definitions only)
Theorem: If a does not divide b2 , then a does not divide b.
Proof by contrapositive method: Restatement will be that If a divides b then a divides b2.
Suppose a divides b, then by definition of divisible there exist a q so that aq = b. Now multiply
both sides by b to get that aqb = bb. We now have that (a)(m) = b2 where m = qb is an integer.
Therefore by definition of divisible, a divides b2. Since we have proved the contrapositive true,
then the original equivalent statement is also true.
Problem 3: Use the Quotient Remainder Theorem solve the following problems.
Solutions: (12pts)
A) Given that x mod 5 = 2 and y mod 5 = 4, find (2x +3y) mod 5.
Solution: By QRT x = 5q + 2 and y = 5k + 4 where q and k are positive integers.
Now consider 2x + 3y = 2(5q + 2) + 3(5k + 4) = 10q + 15k +16 = 5(2q + 3k + 3) + 1.
Therefore by the QRT we have that 2x + 3y = 5p + 1 where p = 2q + 3k + 3 is an integer
and (2x + 3y)mod 5 = 1
B) Determine the representation of 138 in terms of the integers mod 5.
By the QRT we have that 138 = 5(27) + 3 so 38mod 5 = 3.
Problem 4: (10pts) Using definitions and /or theorems, to prove or disprove the following:
Theorem: The product of any two irrational numbers is irrational.
Counter example: ( 3)( 3)  3 where
3 is irrational but 3 is an integer and rational.
Problem 5: (16pts) Prove the following by contradiction.
Theorem: If x is irrational then 2x + 5 is irrational for x  R.
Proof by contradiction: Restatement is that assume that x is irrational and 2x + 5 is rational.
a
By the definition of rational 2 x  5  , for non zero -integers a,b . Now solve for x.
b
a
2x  5 
b
a  5b m
x
 , where m = a  5b and n = 2b are integers
2b
n
By definition of rational we have that x is rational which is a contradiction since x is irrational.
Therefore 2x + 5 is irrational.
Problem 6: Using definitions and/or theorems, to prove or disprove the following:
Theorem: For natural numbers m and n, if the product (m)(n) is even and the sum m + n is
even then m and n are each even.
Proof: (16pts) Given that m + n is even, then by definition of even there exist an integer q so
that m + n =2q so n = 2q – m. We also have that (m)(n) is also even and by definition of even
(m)(n) = 2k for some integer k. so m(2q – m) = 2k and 2mq – m = 2k so m = 2(mq-k) = 2f
Where f is an integer and by definition of even m is even. Now consider n = 2q – m. since the
difference of two even integers is even we have that n is also even.
Problem 7: (16pts) Using one or more of the following previously proved theorems listed
below, prove the following Theorem (but first read directions below. )
Theorem:
If a and b are each odd integers and c > 2 is a prime number, then a 2 +(b+1)2 + c is even.
Proof: Since a is odd we know that a2 = (a)(a) is also odd since the product of two odd integers
is odd. Since b is odd we know that b + 1 is even since consecutive integers have opposite
parity. (b + 1 )2 = (b+1)(b+1) is also even since the product of two even integers is even. The
integer c is prime and therefore odd. The sum of a2 + c is even since the sum of two odd
numbers is always even. Therefore (a2 + c ) + (b + 1)2 is also even, since the sum of two even
integers is even.
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List of previously proved theorems: Use only the theorems needed and quote the ones you
decide to use. Do not try to reprove any of these theorems.
Two consecutive integers have opposite parity
The product of two consecutive integers is even.
The product of two even (odd) integers is even (odd).
The sum of any two odd integers or any two even integers is even.
Any prime number greater than 2 is an odd integer.
The sum of any two integers is an integer.
The sum of an even and odd integer is odd.
The natural number n2 is even if and only if n is even.