Homework #3 - This homework adds up to 21 points
Section1.8
2 – 1 pt Let n be an even number. Than it has the form n=2k. To show that n2 is even substitute the form of n
and square thus: n2= (2k)2= 4k2= 2(2k2) =2i which has the form of an even number therefore n2 is even.
16 – 2 pts (1 pt for each proof direction) First, assume that n is odd, so that n = 2k+1 for some integer k. Now
substitute in the form so then
5n+6 = 5(2k+1)+6 = 10k + 11 = 2(5k + 5) + 1. Hence, 5n + 6 is odd.
To prove the converse, suppose that n is even, so that n = 2k for some integer k.
Then 5n + 6 = 10k + 6 = 2(5k + 3), so 5n + 6 is even.
Hence, n is odd if and only if 5n + 6 is odd.
20 – 1pt No. This line of reasoning shows that if √2𝑥 2 − 1, then we must have x = 1 or x = −1. These are
therefore the only possible solutions, but we have no guarantee that they are solutions, since not all of our steps
were reversible (in particular, squaring both sides). Therefore we must substitute these values back into the
original equation to determine whether they do indeed satisfy it.
Homework assignment for week 5 (Feb 8 th)
1 - 4 pts Use a table to express the values of each of these Boolean functions
a)
b)
c)
d)
a)
F (x,y,z) = z’
F (x,y,z) = x’y + y’z
F (x,yz) = xy’z + (xyz)’
F(x,y,z) = y’(xz + x’z’)
b)
c)
d)
2 - 4 pts Find the sum-of-products and product-of-sums expansions of these Boolean functions:
a)
b)
c)
d)
F(x,y) = x’+ y
F(x,y) = xy’
F(x,y) = 1
F(x,y) = y’
a) F(x,y) = x’ + y
= x’∙1 + y∙1
= x’(y+y’) + y(x+x’)
= x’y + x’y’ + yx + yx’
= x’y + x’y’ + xy + x’y
= x’y + x’y’ + xy
Commutative law
Idempotent law
b) It is already in the product of sums expansion
c) F(x,y) = 1 = x + x’
= x∙1 + x’∙1
= x(y+y’) + x’(y+y’)
= xy+ xy’ +x’y + x’y’
d) F(x,y) = y’ = 1∙y’
= (x+x’)y’
= xy’ + x’y’
3 - 2 pts Find the output of the given circuits:
a)
b)
4 - 6 pts (2 pts for each circuit)
outputs
a) x’ + y
b) (x+y)’x
c) xyz + x’y’z’
Construct circuits from inverters, AND gates, and OR gates to produce these