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Needing some help with Discrete
#1
Needing some help with Discrete
Currently studying for a test and noticed this problem


2.1.23. How many elements does each of these sets have where a and b are distinct elements?

A. P({a, b, {a, b}})
B. P({∅, a, {a}, {{a}}})
C. P(P(∅))

For whatever reason this was not covered in class (It's not in my notes and i take notes in that class religiously) Was wondering if anyone knew how the hell do this, the book we were assigned does a poor job of explaining it, or anything for that matter.
~Lane~
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#2
RE: Needing some help with Discrete
[Image: 931.jpg]
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#3
RE: Needing some help with Discrete
Is P being defined as the set of subsets including the empty set 0 and the set itself?

Then P(0)= {0}

and P(P(0))={0,{0}}

You can take it from here Smile

P.s. I think if a set S has n elements, P(S) always has 2^n, because each can be in a subset or not.
The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.
Psalm 14, KJV revised edition

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#4
RE: Needing some help with Discrete
(April 2, 2014 at 5:07 pm)Alex K Wrote: Is P being defined as the set of subsets including the empty set and the set itself?

I've no clue, that's literally it says..
~Lane~
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#5
RE: Needing some help with Discrete
The empty set is.
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#6
RE: Needing some help with Discrete
(April 2, 2014 at 5:07 pm)Alex K Wrote: Is P being defined as the set of subsets including the empty set 0 and the set itself?

Then P(0)= {0}

and P(P(0))={0,{0}}

You can take it from here Smile

What u will end up getting is opp.... yeah u know me...good song yo!!!
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#7
RE: Needing some help with Discrete
(April 2, 2014 at 4:05 pm)SavageNerdz Wrote: Currently studying for a test and noticed this problem


2.1.23. How many elements does each of these sets have where a and b are distinct elements?

A. P({a, b, {a, b}})
B. P({∅, a, {a}, {{a}}})
C. P(P(∅))

For whatever reason this was not covered in class (It's not in my notes and i take notes in that class religiously) Was wondering if anyone knew how the hell do this, the book we were assigned does a poor job of explaining it, or anything for that matter.

Wikipedia Wrote:Writing A = {1, 2, 3, 4 } means that the elements of the set A are the numbers 1, 2, 3 and 4. Sets of elements of A, for example {1, 2}, are subsets of A.

Sets can themselves be elements. For example consider the set B = {1, 2, {3, 4}}. The elements of B are not 1, 2, 3, and 4. Rather, there are only three elements of B, namely the numbers 1 and 2, and the set {3, 4}.

The elements of a set can be anything. For example, C = { red, green, blue }, is the set whose elements are the colors red, green and blue.

A. P({a, b, {a, b}})
B. P({∅, a, {a}, {{a}}})
C. P(P(∅))


It's been too long, so don't take my word for it, but...

A. one element (the set {a,b,{a,b}}), or three, depending.
B. one element or four, again depending.
C. Not sure. I don't know if the empty set counts as an element; my guess would be, yes, it does.
[Image: extraordinarywoo-sig.jpg]
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#8
RE: Needing some help with Discrete
rasetsu, I think you gotta go 1 level deeper, inside the first bracket.
The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.
Psalm 14, KJV revised edition

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#9
RE: Needing some help with Discrete
The back of the book says this is the answers:

A: 8
B: 16
C: 2

Which is why i'm so confused, it is completely blowing my mind right about now..
~Lane~
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#10
RE: Needing some help with Discrete
That's right, the formula is 2^n:

8=2^3
16=2^4
2=2^(2^0)

Example:The first set contains 3 elements: a, b, {a,b}

You can form the 8 subsets
0={}
{a}
{b}
{a,b}
{{a, b}}
{a,{a,b}}
{b,{a,b}}
{a,b,{a,b}}

Note the difference betw the fourth and the fifth. Thw fourth contains two elements, a and b, the fifth set contains one element, which by coincidence is the fourth set itself again to confuse you.
The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.
Psalm 14, KJV revised edition

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