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# System prepended metadata

title: SS lektion 0

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# SS lektion 0

![](https://i.imgur.com/Xdk5FUs.png)
1. false
2. true
3. true
4. false?
5. maybe. Er empty set et subset af A?
6. false
7. true
8. true
9. false
10. true
11. true
12. false
![](https://i.imgur.com/QJxFN8S.png)
a) A U B = {x,y,z}
b) A ∩ B = {x,y}
c) *P*(B) = {Ø,{x},{y},B}
d) A × B = {(x,x),(x,y),(y,x),(y,y),(z,x),(z,y)}


![](https://i.imgur.com/L2IJnE3.png)
a) A U B = {n \| n ∈ A V n ∈ B}
b) A ∩ B = {n \| n ∈ A ∧ n ∈ B}
c) *P*(A) = {n \| n ∈ A<sup>k</sup>} n is a set with k elements
d) (A * B) ∩ (C * B)= {(x,y) \| x ∈ (A ∩ C) ∧ y ∈ B}
    kan også skrives som {(x,y) \| x ∈ A ∧ x ∈ C ∧ y ∈ B}
![](https://i.imgur.com/0J2BVWl.png)

![](https://i.imgur.com/dZ0e3IF.png)

![](https://i.imgur.com/2XQKDUv.png)

![](https://i.imgur.com/eYwrsg1.png)

![](https://i.imgur.com/J8s7D2h.png)
-   m, n ∈ Z. If m + n ≥ 19, then m ≥ 10 or n ≥ 10.
    **Proof by contradiction:**
    Then for the sake of the contradiction m, n < 10. By this assumption: m + n < 20.
    m + n = 19
    m + n -10 = 9 (substract 10 on both sides)
    Then n-10 is negative because n < 10
    Therefore m > 9. There is no integer that satisfies m > 9     and m < 10. Thus it is a contradiction.

![](https://i.imgur.com/Jw9aigc.png)

![](https://i.imgur.com/H9FXis1.png)
**Base case:**
Let P(n) be the property "there exists an integer m such that "4<sup>n</sup> + 14 = 6m"
Base case: P(1). 4<sup>1</sup> + 14 = 6 * 3.

**Inductive step:**
For i ≥ 1 assume P(i) is true. Then 4<sup>i</sup> + 14 = 6m, by rearranging we have 4<sup>i</sup> = 6m - 14 (inductive hypothesis). Then we shall show that p(i+1) follows:
4<sup>i+1</sup> + 14 = 4(4<sup>i</sup>) + 14
                     = 4(6m-14)+14 (inserting inductive hypothesis)
                     = 24m-42
                     = 6(4m-7)
m'=4m-7, it is the integer showing that P(i+1) holds.
![](https://i.imgur.com/pWsCM8f.png)




