#image("Opgaver-numre.png")
= Opgave 1.7-3
#image("opgave3-1.7.png")
For an even number you know that $a=2k$.
Thus you can say $a²=(2k)²=4k²=2(2k²)$

Now you have $2(2k²)$ which is the same form as the formula for an even number, thus it's proven.

= Opgave 1.7-11
#image("opgave11.png")
If you for example take the irrational number $sqrt(2)$. The product of that is $sqrt(2)²=2$. This disproves that the product of two irrational numbers are irrational.
= Opgave 1.7-37
#image("opgave37.png")
Because there is not a biimplification between step 1 and 2. When you take the square root of something, you can't nescecarily go back, because you have to go back to a $plus.minus$. You can prove this by x=6. For step 2, this is a correct x to solve the equation, but not for step 1. When you have $sqrt(x+3)$, and you take the square too, you have $(plus.minus sqrt(x+3))²$.
= Opgave 1.7-41
#image("opgave41.png")
The equal to is easy to prove. If you have only 1 number or take the average of $a_n$ of same value, the average will then be the same value.

To prove the theory, you can say the opposite, which is that all the values of a is less than the sum.
This is provably false. The equation to get the average is $A=(a_1+a_2+a_3+...+a_n)/n$. We can get this from it: $A*n=a_1+a_2+a_3+...+a_n$.

If we say that all numbers of _a_ is less than the average, then all of them added together must be less than the average times the amount of elements: $a_1+a_2+a_3+...+a_n<A*n$

Now we have a contradiction. We know that when you prove the opposite, then the original must be true, and thust the statement is true.

I used proof by contradiction

= Opgave 2.1-11
#image("opgave-2.1-11.png")
a) False. Empty set has no elements

b) False. Only an empty set contains nothing

c) False. No elements in empty set

d) True. An empty subset is a subset of everything.

e) False.

f) False. True {0} is a subset of {0}, but they are equal to each other, so its false

g) True.
= Opgave 2.1-21
#image("Opgave 2.1-21.png")
a) 1 (contains 1 element)

b) 1 (contains 1 element: a list of another element)

c) 2 (contains 2 elements: a and a list)

d) 3 (contains 3 elements: a, and 2 lists)


= Opgave 2.1-27
#image("opgave 2.1-27.png")
For $cal(P)(A)$ to be a subset of $cal(P)(B)$, then A must have been a subset of B to begin with. If A is a subset of B, then the elements of the powerset of A must be contained in the powerset of B (alongside many more). If A was not a subset of B (containing an element not in B), then its powerset would also have elements not in the powerset of B and thus not be a subset.

= Opgave 2.1-37
#image("Opgave 2.1-37.png")
Its size would be $|A|*|B|$. When you calculate $A times B$ you calculate every "variation" of the elements of both, so for $a_1$ you pair it with every element of B until $a_n$. Thus it's their lengths times eachother