Summary Assignment 1 Full Attempt 100% COS1501

UNISA  2023  COS1501-23-Y  Welcome Message  Assessment 1

QUIZ



COS1501 Assignment 1 Memo Review
Started on Tuesday, 2 May 2023, 6:42 PM
State Finished
Completed on Tuesday, 2 May 2023, 6:50 PM
Time taken 7 mins 57 secs
Marks 15.00/15.00
Grade 100.00 out of 100.00


Question 1 Which one of the following alternatives is FALSE regarding the number sets Z, Z+, Z≥, Q
Correct and R?
Mark 1.00 out of
1.00
a. Z≥ ⊆ Z
b. Z+ ⊆ Z≥
c. R ⊆ Q 

d. Z+ ⊆ R



The different number sets are described in Chapter 1 and 2.
Z = {... -4, -3, -2, -1, 0, 1, 2, 3, 4, ...} Z≥ = {0, 1, 2, 3, 4, ...} Z+ = {1, 2, 3, 4, ...}
From the diagram below it is clear that all elements in Z+ is also in the set Z≥, and all
elements in Z≥ is also in the set Z etc.




We can therefore say that Z+⊆ Z≥ ⊆ Z ⊆ Q ⊆ R.




The correct answer is: R ⊆ Q




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, Question 2 The set of all non-negative integers x less than 16 such that x2 is an even integer can
Correct be described as the set:
Mark 1.00 out of (Note: The required set must include as elements all non-negative integers x such
1.00 that all the requirements for the set are met.)


a. {x| x ∈ Z≥, x < 
16, x2 = 2k The set {x | x Î Z≥, x < 16, x2 = 2k for some k Î Z} is the set
for some k ∈ of all non-negative integers x less than 16 such that x2 is
Z} an even number.
All the requirements x Î Z≥, x < 16, x2 = 2k for some k Î Z
must hold for an integer x to qualify to belong to this set.
Case 1: If x = 0 then x2 = 0 and 0 Î Z≥, 0 < 16, 02 = 2(0) for
some k = 0 ∈ Z.
Case 2: If x = 2 then x2 = 4 and 2 Î Z≥, 2 < 16, 22 = 2(2) for
some k = 2 ∈ Z.
And so we can go on to see that for all non-negative
integers x = 0, 2, 4, 6, 8, 10, 12, 14 it will be the case that
x2 = 2k, k ∈ Z, and also note that x < 16.
If x = 1, 3, 5, 7, 9, 11, 13, 15 then x does not qualify to
belong to the required set since x is an odd number and
thus x2 is also an odd number.
Refer to study guide, pp 3, 7, 11, 35 – 39.

b. {x| x ∈ Z≥, x < 16, x2 = 2k for some k ∈ Z+}
c. {0, 2, 4, 16, 36, 64, 100, 144, 196}
d. {0, 2, 4}




Refer to study guide, pp 3, 7, 11, 35 – 39.
The correct answer is: {x| x ∈ Z≥, x < 16, x2 = 2k for some k ∈ Z}




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