Mathematics, 24.01.2020 21:31 live4dramaoy0yf9

Produce an infinite collection of sets a1, with the property that every ai has an infinite number pf elements , ai inter aj = empty set for all i, j different.

and uinfinitii=1ai =n

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Answer from: carlosgc19

$A_n = [2^{n-1} , 3(2^{n-1}), 5(2^{n-1}), 7(2^{n-1}),....]$

Step-by-step explanation:

For this case we need to produce an infinite collection of sets A_a, A_2, A_3,.... with the property that every A_i has an infinite number of elements with $A_i \cap A_j , i\neq j$ and $\cup_{i=1}^{\infty} A_i = N$

So for this case we need to create a set A who satisfy 3 conditions (Infinite number of elements, Disjoint and Union represent the natural numbers)

So if we define the nth term for the set A:

$A_n = [2^{n-1} , 3(2^{n-1}), 5(2^{n-1}), 7(2^{n-1}),....]$

We see that the set A represent all the odd multiplies of $2^{n-1}$ and if we check the properties we have this:

Disjoint

If we select A_n , A_m with $n\neq m$ and we can assume for example that and if we have an element a in the intersection of the sets $a \in A_n \cap A_m$ , so then needs to exists some odd numbers k and l such that

$a = 2^{n-1} k = 2^{m-1}l$

And since we assume that then we have that $n\leq m-1$ and we can write:

$2^{m-1} =2^m 2^i , i\geq 0$

And then $2^{n-1} k= 2^n 2^i l$ and if we divide by $2^{n-1}$ we got:

k = 2 2^i l so then k is not odd since the last statement contradicts this. So then we can conclude that $A_n \cap A_m = \emptyset$

Union

For this case we need to show that $\cup_{i=1}^{\infty} A_i = N$

Since each element A_n is a subset of the natural numbers then the unision of the sets represent N

For the other side of the explanation if we assume that $a\in N$ we can write $a= 2^{n-1}k$ for any $n\in N$ and k odd, and by this $a\in A_n$ and we chaek the property.

Infinite condition

For this case $A_n = [2^{n-1} , 3(2^{n-1}), 5(2^{n-1}), 7(2^{n-1}),....]$

is an infinite set since we don't have a limit for n so then we have infinite elements for this case.

And since all the properties are satisfied we end the problem.

Answer from: Quest

the law of definite proportions states samples of a compound will always contain the same proportion of elements by mass. in other words, a given compound will always contain the same elements in the same proportions by mass.the law of multiple proportions states when two elements combine with each other to form more than one compound, then the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

Answer from: Quest

The point (5, 10) will work for the rule y=x+5.y= 10 x= 510=5+510=10therefore the ordered pair (5, 10) works for the rule

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