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snow_tiger [21]
3 years ago
14

Prove the set of subsets of a finite set has cardinality 2^x

Mathematics
1 answer:
Nata [24]3 years ago
8 0

Let's take the simple case of a set with four elements: the letters a-d

\{a,b,c,d\}

Two subsets that this - and any other - set contains are the empty set ∅ and the set itself. Now, if we wanted, we could construct the rest of the subsets by picking elements from the original set at random - {a, b, c}, {a, c}, and {c, d} to name a few - but this process is incredibly inefficient, and there's a good chance you'll miss a few subsets this way.

There's a part in that last paragraph that's extremely important: we're <em>picking</em> elements from the original set to put in our subsets, and this selection process boils down to a single yes or no question: <em>do we want to add this element to our subset? </em>This is where that 2 emerges in the original question - we're asking a question with 2 possible outcomes, and we're asking it x times, where x is the number of elements in our set.

For instance, with the set {a, b, c, d}, constructing subsets consists of four questions:

- Should we add a to the subset? Yes/No

- Should we add b? Yes/No

- Should we add c? Yes/No

- Should we add d? Yes/No

The space of possible outcomes, and consequently possible subsets, these questions produce is the same as the space of possible outcomes for 4 yes-or-no questions: 2^4=16

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Find the limit
Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

\large\underline{\sf{Solution-}}

Given expression is

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

On substituting directly x = 1, we get,

\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}

\rm \: = \sf \: \: - \infty \: - \: \infty

which is indeterminant form.

Consider again,

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

can be rewritten as

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]

\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]

\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}

\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}

\rm \: = \: \sf \boxed{2}

Hence,

\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}

\rule{190pt}{2pt}

7 0
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Answer:

see below

Step-by-step explanation:

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FOIL

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Add together

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