MHB How Many Elements Are Found in Each Mathematical Set?

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SUMMARY

The discussion centers on determining the number of elements in two mathematical sets. The first set, defined as the union of {1,2,3,4,5}, {1,2,3,4,5,6}, ..., {1,2,3,...,9000}, contains 9000 elements, as it includes all integers from 1 to 9000. The second set, defined by the condition $$1 PREREQUISITES

  • Understanding of set theory and unions
  • Familiarity with modular arithmetic
  • Knowledge of mathematical notation for inequalities
  • Basic concepts of congruence relations
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  • Study set theory, focusing on unions and intersections
  • Learn about modular arithmetic and its applications
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lemonthree
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Question: How many elements are in each set?

For the first set, I think it's 8995 because the set is the union of {1,2,3,4,5},{1,2,3,4,5,6},...{1,2,3,...9000}. So 9000 - 5 = 8995.

For the second set, I'm not too sure about counting the elements in the set. Since $$1<x≤i$$, I can't think of any x mod i = 2.
For example, I know 5 mod 3 = 2, but 5 > 3 and in this case it wants i to be greater or equal to x...any hints please?
counting-sets.png
 
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There are, of course, 9000 integers from 1to 9000. Why are you subtracting 5? Which integers are missing?

A number, n, is congruent to 2 (mod i) if n= i+ 2. Every number, except 1 and 2, is equal to i+ 2 for some i.
 
lemonthree said:
Question: How many elements are in each set?

For the first set, I think it's 8995 because the set is the union of {1,2,3,4,5},{1,2,3,4,5,6},...{1,2,3,...9000}. So 9000 - 5 = 8995.
I think it's 8996, because you need to count both endpoints.

lemonthree said:
For the second set, I'm not too sure about counting the elements in the set. Since $$1<x≤i$$, I can't think of any x mod i = 2.
For example, I know 5 mod 3 = 2, but 5 > 3 and in this case it wants i to be greater or equal to x...any hints please?
If $i=1$ then the set $\{x\ |\ x \text{ is an integer and } 1<x\leqslant i \text{ and }x=2\pmod i\}$ is the empty set. For all other values of $i$ that set just consists of $x=2$. So your second set is $\emptyset\cup\{2\}$. It therefore contains two elements.
 
Both of you are quite right;

For the first question, there are 9000 elements. @Country Boy How do you know that there are 9000 elements though? Doesn't that symbol represent the union of indexed collection from i = 5 to i = 9000? I see it to be similar to the summation notation but I guess that's where I'm wrong.

For the second question, there is 1 element, i.e. {2}, so @Opalg you are right. We take ∅∪{2} to be equal to {2}. Thank you for the explanation, I realized I could view it as 2 = 0*i + 2, for various i values until infinity, which made sense for {2} to be the only element.
 
If $A\subseteq B$ then $A\cup B= B$. These sets are "nested" so the union is just the largest set.
 
lemonthree said:
Both of you are quite right;

For the first question, there are 9000 elements. @Country Boy How do you know that there are 9000 elements though? Doesn't that symbol represent the union of indexed collection from i = 5 to i = 9000? I see it to be similar to the summation notation but I guess that's where I'm wrong.

For the second question, there is 1 element, i.e. {2}, so @Opalg you are right. We take ∅∪{2} to be equal to {2}. Thank you for the explanation, I realized I could view it as 2 = 0*i + 2, for various i values until infinity, which made sense for {2} to be the only element.
Yes, you are correct. In both cases I was thinking in terms of a set of sets rather than a union of sets.
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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