CSCI-2041: Advanced Programming Principles 2024 –
2025 CSCI-2041 Summer Exam Quiz 2 Review Questions
with Verified Solutions | 100% Pass Guaranteed | Graded
A+ |
Administrator
[COMPANY NAME] [Company address]
, Name: ID#: X500: @umn.edu A
CS 2011: Practice Quiz 2 SOLUTION
Summer 2018
University of Minnesota
Quiz period: 15 minutes
Points available: 20
Problem 1 (10 pts): Show that for real numbers x, y x ≤ x+y
OR y ≤ x+y
. For full credit, include
2 2
one of the following in your proof.
• Clearly defined cases which are proved independently
• OR Proper use of the notion of “Without loss of generality”.
SOLUTION: Without loss of generality, assume that x ≤ y. This will lead to the first inequality, x ≤ x+y
2
holding while the reasoning would be identical with the opposite assumption.
Re-arranging the original inequality gives
x+
y
x≤ (1)
2
2x ≤ x + y (2)
x≤ y (3)
This last inequality is given as a fact in the assumption which we know to be true proving the case.
Alternatively, one might transform the fact x < y into the given equality.
Problem 2 (10 pts): Show that following equivalence involving sets A, B holds.
A−B=A∩B
You may use any method to do show the equivalence so long as you reasoning is explained.
SOLUTION: Below is a series of transformations in set builder notation the equivalence.
A − B = {x|x ∈ A ∧ x /∈ Def. of set difference (4)
B}
= {x|x ∈ A ∧ x ∈ B} Def. of set complement (5)
= {x|x ∈ (A ∩ B)} Def. of set intersection (6)
=A∩B Simplification (7)
2025 CSCI-2041 Summer Exam Quiz 2 Review Questions
with Verified Solutions | 100% Pass Guaranteed | Graded
A+ |
Administrator
[COMPANY NAME] [Company address]
, Name: ID#: X500: @umn.edu A
CS 2011: Practice Quiz 2 SOLUTION
Summer 2018
University of Minnesota
Quiz period: 15 minutes
Points available: 20
Problem 1 (10 pts): Show that for real numbers x, y x ≤ x+y
OR y ≤ x+y
. For full credit, include
2 2
one of the following in your proof.
• Clearly defined cases which are proved independently
• OR Proper use of the notion of “Without loss of generality”.
SOLUTION: Without loss of generality, assume that x ≤ y. This will lead to the first inequality, x ≤ x+y
2
holding while the reasoning would be identical with the opposite assumption.
Re-arranging the original inequality gives
x+
y
x≤ (1)
2
2x ≤ x + y (2)
x≤ y (3)
This last inequality is given as a fact in the assumption which we know to be true proving the case.
Alternatively, one might transform the fact x < y into the given equality.
Problem 2 (10 pts): Show that following equivalence involving sets A, B holds.
A−B=A∩B
You may use any method to do show the equivalence so long as you reasoning is explained.
SOLUTION: Below is a series of transformations in set builder notation the equivalence.
A − B = {x|x ∈ A ∧ x /∈ Def. of set difference (4)
B}
= {x|x ∈ A ∧ x ∈ B} Def. of set complement (5)
= {x|x ∈ (A ∩ B)} Def. of set intersection (6)
=A∩B Simplification (7)
