Group 4 Project
James Earthman, Gregg Greenleaf, Zach
Harvey, Zac Leids
2.5 #21
21. Give the output signals S and T for the circuit in the right
column if the input signals P, Q, and R are as specified.
Note that this is not the circuit for a full-adder.
a. P = 1, Q = 1, R = 1
b. P = 0, Q = 1, R = 0
c. P = 1, Q = 0, R = 1
2.5 #21 Solution
a. S = 0, T = 1
b. S = 1, T = 0
c. S = 0, T = 0
A Half-Adder takes two binary digits and adds them together. The output for a
half-adder is given in two parts, the carry and the sum. The carry is returned
true when both of the inputs are true or 1, this is like carrying a digit in decimal
addition. The sum is returned true when one of the inputs is true but not both.
Both the carry and the sum are returned false if both inputs are false. This
circuit also uses an AND gate. An AND gate only returns a true value when
both inputs are true, it returns false in any other case.
Gottfried Leibniz
Gottfried Leibniz was the mathematician who invented
the Binary number system. Leibniz was a German
academic that excelled in areas of mathematics and
philosophy. He is infamously known for inventing
Calculus independant of Isaac Newton. Newton later
claimed that Leibniz had stolen his work. Although he
was an amazing mathematician, he also wrote on politics
and law. Binary number theorem is the foundation for
modern computing.
2.5 #22
22. Add 11111111 + 1 and convert the result to
decimal notation,
to verify that 11111111 = (2∨8 − 1)10.
2.5 #22 Solution
11111111
+
1
= 100000000
100000000 -> Decimal = 256 or 2∨8
When adding binary numbers, 1 is carried when both digits are 1 this leaves behind a sum of 0. In this
problem, the a 1 is carried all the way to the ninth digit leaving behind 0’s is each column. When
converting the binary number to decimal when can assign a binary value to each digit via 2∨Digit N,
then take the sum of those values and calculate our decimal number. For this problem, we see that
there is a 1 in the 8th digit giving us a value of 256, since none of the other digits are true our answer
is just 256.
2.5 #47
Part 1: Convert 61502 to decimal notation.
Part 2: Convert 20763 to decimal notation.
2.5 #47 Solution
Part 1:
61502 = 6 * 84 + 1 * 83 + 5 * 82 + 0 * 81 + 2 * 80 = 25410
Part 2:
20763 = 2 * 84 + 0 * 83 + 7 * 82 + 6 * 81 + 3 * 80 = 8691
3.3 #56
Let P(x) and Q(x) be predicates and suppose D is the domain
of x. In 55–58, for the statement forms in each pair, determine
whether (a) they have the same truth value for every choice of
P(x), Q(x), and D, or (b) there is a choice of P(x), Q(x), and
D for which they have opposite truth values.
∃ x ∈ D, (P(x) ∨ Q(x)) and (∃ x ∈ D, (P(x)) ∨ (∃ x ∈ D, Q(x))
3.3 #56 Solution
a) No matter what the domain D or the predicates P(x) and Q(x) are, the given
statements have the same truth value. If the statement “∃ x ∈ D, (P(x) ∨ Q
(x))” is true, then P(x) ∨ Q(x) is true for the x chosen in D, which implies that
both P(x) and Q(x) are true for the x chosen in D. But then P(x) is true for the x
chosen in D, and also Q(x) is true for the x chosen in D. So the statement “(∃ x
∈ D, (P(x)) ∨ (∃ x ∈ D, Q(x))” is true. Conversely, if the statement “(∃ x ∈
D, (P(x)) ∨ (∃ x ∈ D, Q(x))” is true, then P(x) is true for the x chosen in D, and
also Q(x) is true for the x chosen in D. This implies that both P(x) and Q(x) are
true for the x chosen in D, and so P(x) ∨ Q(x) is true for the x chosen in D.
Hence the statement “∃ x ∈ D, (P(x) ∨ Q(x))” is true.
3.3 #58
Let P(x) and Q(x) be predicates and suppose D is the domain
of x. In 55–58, for the statement forms in each pair, determine
whether (a) they have the same truth value for every choice of
P(x), Q(x), and D, or (b) there is a choice of P(x), Q(x), and
D for which they have opposite truth values.
∃x ∈ D, (P(x) ∨ Q(x)), and (∃x ∈ D, P(x)) ∨ (∃x ∈ D, Q(x))
3.3 #58 Solution
a) If the statement “∃ x ∈ D, (P(x) ∨ Q(x))” is true, then P(x) ∨ Q(x) is true for
the x chosen in D, since only P(x) or Q(x) has to be true for the chosen x in D.
So the statement “(∃ x ∈ D, (P(x)) ∨ (∃ x ∈ D, Q(x))” must be true.
Conversely, if the statement “(∃ x ∈ D, (P(x)) ∨ (∃ x ∈ D, Q(x))” is true, then
P(x) is true for the x chosen in D, or Q(x) is true for the x chosen in D. This
implies that P(x) or Q(x) are true for the x chosen in D, and so P(x) ∨ Q(x) is
true for the x chosen in D. Hence the statement “∃ x ∈ D, (P(x) ∨ Q(x))” is
true.
References
http://en.wikipedia.org/wiki/List_of_mathematical_symbols
http://academic.evergreen.edu/projects/biophysics/technotes/misc/bin_math.
htm
http://www.wikihow.com/Convert-from-Binary-to-Decimal
http://simple.wikipedia.org/wiki/Gottfried_Leibniz