Let me rephrase it. You mean take a bounded subset of real numbers, S, and find a subset of all the upperbounds of S, say D, such that sup S is not in D?
If I get you right, then yes.
Take D := {a : a = sup s + n, n is natural and n < 4} so the first element is sup s + 1 > sup s, and the remaining two terms are even larger than the first one.
But I think I got you wrong, go through the Completeness Axiom.
That is for any two set A and B such that for all a in A, b in B, a <= b, and they share a maximum of one element, then there exist at least one number x such that a <= x <= b
In particular, A have a supremum, sup A <= x and B have an infimum inf B > = x
sup A <= inf B if they are equal then they must be x.
Yes, that is true.
The proof is by contradiction: assume there is a finite number of prime numbers and get a contradiction by requiring a prime that is not one of the finite number of primes. Suppose there are only a finite number of prime numbers. Then there are n of them.; and they can all be listed as: p1, p2, ..., pn in order with there being no possible primes between p(r) and p(r+1) for all 0 < r < n. Consider the number m = p1 × p2 × ... × pn + 1 It is not divisible by any prime p1, p2, ..., pn as there is a remainder of 1. Thus either m is a prime number itself or there is some other prime p (greater than pn) which divides into m. Thus there is a prime which is not in the list p1, p2, ..., pn. But the list p1, p2, ..., pn is supposed to contain all the prime numbers. Thus the assumption that there is a finite number of primes is false; ie there are an infinite number of primes. QED.
Actually a stronger statement can be made:A group G is finite if and only if the number of its subgroups is finiteLet G be a group. If G is finite there is only a finite number of subsets of G, so clearlya finite number of subgroups.Now suppose G is infinite , let'ssuppose one element has infinite order. The this element generates an infinite cyclicgroup which in turn contains infinitely many subgroups.Now suppose all the subgroups have finite order Take some element of G and let it generate a finite group H. Now take another element of G not in H and let it generate a finite group I. Keep doing this by next picking an element of G not H or I. You can continue this way.
Each integer has a finite number of factors and an infinite number of multiples.
GCF(12, 224, 36, 48) = 12. Since it is a finite sequence, it can be continued in infinitely many different ways and so there are rules such that any integer (or fraction) can belong to the pattern.
Finite means to end. So finite math is possible math that has to do with numbers ending; no irrational(unending) numbers. Ex: 2+2=4
All real numbers are finite. Infinity is not a number.All real numbers are finite. Infinity is not a number.All real numbers are finite. Infinity is not a number.All real numbers are finite. Infinity is not a number.
Finite.
Finite, no.
Finite
There are a finite number of apartments. Finite numbers may be large or small. There are a finite number of states. The number of molds in my fridge is not exactly finite.
Yes, finite numbers are always countable.
They are numbers that terminate.
noo
Finite math is a branch of mathematics that focuses on mathematical concepts that have finite, specific values or elements. It typically includes topics such as logic, set theory, probability, statistics, and linear algebra, which are applied to real-world problems with a limited number of possibilities. Finite math is often used in business, social sciences, and computer science to model practical situations.
No; there are infinitely many rational numbers.
A finite set is a set that has numbers you can count. Its not like infinite with no end it has an end.