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Home / Science, Engineering & Maths / Maths / Precalculus: the Mathematics of Numbers, Functions and Equations / Integers in practice

Integers in practice

Solving exercises about integers
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10.8
Hello, and welcome to this step in practice. Today we are concerned about integers. In exercise 1, we are asked to list two sets. The first set A is the set of elements 7 times n to the square when n is an integer varying between minus 5 and 5.
37.2
So if we begin with n equal to minus 5, we get 7 times minus 5 to the square, 7 times minus 4 to the square, and so on. But now, realize that when n is an integer, well minus n to the square is n to the square. So for instance, 7 times minus 5 to the square equals 7 times 5 to the square. So it is enough to consider the integers to vary from 0 to 5. And so the set is 7 times 0 to the square, 7 times 1 the square, 7 times 2 to the square, 7 times 3 to the square, 7 times 4 to the square, and finally 7 times 5 to the square.
85.4
The set B, well, equals the set of numbers of the form m times m plus 1 over 2, where m is an integer– actually a natural number– varying between 1 and 5. So we’ve got 5 elements, starting with m equal to 1, we get 1 times 2 over 2, 2 times 3 over 2, 3 times 4 over 2, 4 times 5 over 2, 5 times 6 over 2, which gives 1, 3, 6, 10, and 15.
139.8
Now, we want to find– this is the first question. In the second question, we are asked to find the intersection of A with B. Well one method is to calculate 7 times 2 to the square and so on. But notice that every element of a is a multiple of 7.
162.4
Elements of a are multiple of 7. And look– in B, there are no multiples of 7. No number belonging to B can be divided by 7. So no multiples of 7 in B. Therefore, the intersection of A with B is the empty set. And as a consequence, the union of A with B, well, in this union, there are all the elements of A and all the elements of B. So 7 times 0 to the square, 7 times 1 to the square, 7 times 5 to the square, and then the elements of B– 1, 3, 6, 10, 15.
218.3
How about A minus C? Well, C is the set of elements of 7 times n to the square, where n varies between 0 and 5, but it’s an old friend. It’s the set A. And therefore, A minus C is A minus A, that is the empty set. And this ends exercise 1.
246.9
In exercise 2, we are given three non-empty sets. A, B, and C. And the set A is used in order to perform a test with the sets B and C. Actually, we know that the intersection of B with A is the intersection of C with A, and the union of B with the set A is the union of the set C with the set A. We are asked to prove that B equals C. Now, in order to prove that two sets are equal– for instance that B equals C– it is enough and necessary to prove that every element of B belongs to C, and conversely, every element of C belongs to B.
300.9
Now, let us take an element x, belonging to the set B. Now, we can consider two cases. Well, if B– if x belongs also to A, then x belongs to the intersection of B with A. But we know that the intersection of B with A is the intersection of C with A. And thus we get that x belongs to C. In the other case, x does not belong to A. Well, how can we take advantage of this information? You already used the first, the fact that the intersections are equal. We did not use the second one, information about the union. But we can use it here, in this step.
362.5
Because you know if x does not belong to A, well, for sure it belongs to B union A. Because it belongs to B. But it does not belong to A. So minus A. However, the union of B with A is the union of C with A. So it belongs to the union of C with A, but not to A. Well, but if an element belongs to the union of C with A, and does not belong to A, well it must belong to C. And so x belongs to C. You see in any case, starting from x belonging to B, we reach the conclusion that x belongs to C, and this proves that B is contained in C.
414
Now, if we reverse the role of the sets B and C, we are able to prove in the same way that C is included in B. And therefore the two sets are equal. And this ends exercise 2.

The following exercises are solved in this step.

We invite you to try to solve them before watching the video. In any case, you will find below a PDF file at the bottom of the step with the solutions. The symbol (in) means “belongs”, so that (xin mathbb R) means that (x) belongs to (mathbb R).

Exercise 1.

Consider the sets ( (mathbb Z=lbrace …,-3,-2,-1,0,1,2,3,…rbrace))

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Precalculus: the Mathematics of Numbers, Functions and Equations

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[ A =leftlbrace 7n^2: nin mathbb{Z},-5leq nleq 5rightrbrace,] [B =leftlbrace dfrac{m(m+1)}2: min mathbb{Z},1leq mleq 5rightrbrace,] [C =leftlbrace 7n^2: nin mathbb{Z}, 0leq nleq 5rightrbrace. ]

1) List all the elements in (A) and in (B).

2) Determine (Acap B) (the intersection of (A) with (B) , (Acup B) (the union of (A) with (B)), (Asetminus C) (the elements that belong to (A) but not to (C)).

Exercise 2.

Let (A), (B) and (C) be three non-empty subsets of a set (X) such that [Bcap A=Ccap A qquad mbox{and} qquad Bcup A=Ccup A.]

Show that (B = C).

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This content is taken from University of Padova online course
Precalculus: the Mathematics of Numbers, Functions and Equations
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