Let the sets S = {(n, m) ∈ N × N: n + m = 10} and π1 (S) = {n:
there exists m ∈ N such that (n, m) ∈ S}. (Note that π1 (S) ⊆ N.)
(a) Show that 8 ∈ π1 (S) and give an example of an element x
such that x ∈ N ∖ π1 (S).
(b) Show that π1 (S) = N≤10. (Remember that N≤10 is defined
as {n ∈ N: n ≤ 10}.)
(c) Show that, given the sets A, B ⊆ N, S ≠ A × B. (Help:
can reason by contradiction.)
To answer these questions, we need to understand the definitions of the sets and functions involved. Let's break it down step by step.
(a) To show that 8 ∈ π1(S), we need to find an m in N such that (8, m) ∈ S.
The set S is defined as {(n, m) ∈ N × N: n + m = 10}. This means that for any (n, m) in S, n + m must equal 10.
Let's check if (8, 2) satisfies this condition. Here, n = 8 and m = 2. Adding them together gives 10, which means (8, 2) ∈ S. Therefore, 8 ∈ π1(S).
To find an element x that belongs to N ∖ π1(S) (N set minus π1(S) set), we need to find a natural number x that is not in the set π1(S).
Since π1(S) is the set of all possible n-values in S, we need to find an n that does not have a corresponding m such that (n, m) ∈ S.
For example, let's consider n = 3. If we try all possible m-values, we see that there is no m such that (3, m) ∈ S because 3 + m ≠ 10 for all m ∈ N. Therefore, 3 ∈ N ∖ π1(S).
(b) To show that π1(S) = N≤10, we need to prove two things:
1. π1(S) ⊆ N≤10: This means that every element in π1(S) is also in N≤10.
Since S is defined as {(n, m) ∈ N × N: n + m = 10}, n must be a natural number and m must be a natural number. Since n + m = 10, it means that n and m cannot be greater than 10. Therefore, all elements in π1(S) are in N≤10.
2. N≤10 ⊆ π1(S): This means that every element in N≤10 is also in π1(S).
Let's take any element n from N≤10. To prove that it is in π1(S), we need to find an m such that (n, m) ∈ S.
Since n is in N≤10, it means that n ≤ 10. We can choose m = 10 - n. Now, (n, m) = (n, 10 - n), and n + (10 - n) = 10, satisfying the condition for S. Therefore, every element in N≤10 is in π1(S).
Since we proved both π1(S) ⊆ N≤10 and N≤10 ⊆ π1(S), we can conclude that π1(S) = N≤10.
(c) To prove that S ≠ A × B, let's assume that S = A × B. We will try to find a contradiction.
By definition, A × B is the set of all possible (a, b) pairs where a is in set A and b is in set B.
If S = A × B, it means that every pair (n, m) in S must also be in A × B. Therefore, for any (n, m) in S, n must be in A and m must be in B.
However, S is defined as {(n, m) ∈ N × N: n + m = 10}. This means that n and m can be any natural numbers that satisfy n + m = 10, regardless of whether n is in set A or m is in set B.
To find a contradiction, let's consider an example:
Let A = {1, 2, 3} and B = {4, 5, 6}. The possible pairs from A × B are {(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}.
Now, let's check if all these pairs are in S. If we add the numbers in each pair, we can see that none of them equal 10. For example, (1, 4) has a sum of 5, which is not equal to 10.
This contradicts our assumption that S = A × B. Therefore, we can conclude that S ≠ A × B.
By reasoning by contradiction, we have shown that given the sets A, B ⊆ N, S ≠ A × B.