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We prove that if an increasing sequence {an} is bounded above, then it is convergent and the limit is the sup {an }

Now we use the least upper bound property of real numbers to say that sup {an } exists and we call it something, say S. We can say this because sup {an } is not empty and by our assumption is it bounded above so it has a LUB.

Now for all natural numbers N we look at aN such that for all E, or epsilon greater than 0, we have aN > S-epsilon. This must be true, because if it were not the that number would be an upper bound which contradicts that S is the least upper bound.

Now since {an} is increasing for all n greater than N we have |S-an|

Q: How do you prove that if a real sequence is bounded and monotone it converges?

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No. You can always "cheat" to prove this by simply giving the function's domain a bound.Ex: f: [0,1] --> RI simply defined the function to have a bounded domain from 0 to 1 mapping to the codomain of the set of real numbers. The function itself can be almost anything, periodic or not.Another way to "cheat" is to simply recognize that all functions having a domain of R are bounded functions, by definition, in the complex plane, C.(Technically, you would say a non-compact Hermitian symmetric space has a bounded domain in a complex vector space.) Obviously, those functions include non-periodic functions as well.

By "the nth term" of a sequence we mean an expression that will allow us to calculate the term that is in the nth position of the sequence. For example consider the sequence 2, 4, 6, 8, 10,... The pattern is easy to see. # The first term is two. # The second term is two times two. # The third term is two times three. # The fourth term is two times four. # The tenth term is two times ten. # the nineteenth term is two times nineteen. # The nth term is two times n. In this sequence the nth term is 2n.

if EB=EC and AB=DC prove <A=<D

What have we got to prove? Whether we have to prove a triangle as an Isoseles triangle or prove a property of an isoseles triangle. Hey, do u go to ALHS, i had that same problem on my test today. Greenehornet15@yahoo.com

prove- show someone that you can do it, or show someone somthing that they dont believe.

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If the sequence (n) converges to a limit L then, by definition, for any eps>0 there exists a number N such |n-L|N. However if eps=0.5 then whatever value of N we chose we find that whenever n>max{N,L}+1, |n-L|=n-L>1>eps. Proving the first statement false by contradiction.

You can use the comparison test. Since the convergent sequence divided by n is less that the convergent sequence, it must converge.

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It has not yet been proven whether any arbitrary sequence of digits appears somewhere in the decimal expansion of pi.

The limits on an as n goes to infinity is aThen for some epsilon greater than 0, chose N such that for n>Nwe have |an-a| < epsilon.Now if m and n are > N we have |an-am|=|(am -a)-(an -a)|< or= |am -an | which is < or equal to 2 epsilor so the sequence is Cauchy.

No. You can always "cheat" to prove this by simply giving the function's domain a bound.Ex: f: [0,1] --> RI simply defined the function to have a bounded domain from 0 to 1 mapping to the codomain of the set of real numbers. The function itself can be almost anything, periodic or not.Another way to "cheat" is to simply recognize that all functions having a domain of R are bounded functions, by definition, in the complex plane, C.(Technically, you would say a non-compact Hermitian symmetric space has a bounded domain in a complex vector space.) Obviously, those functions include non-periodic functions as well.

There is more than one way to prove a given mathematical proposition. If the sequence of reasoning is valid, then the proof is correct. That is all that is required.

By "the nth term" of a sequence we mean an expression that will allow us to calculate the term that is in the nth position of the sequence. For example consider the sequence 2, 4, 6, 8, 10,... The pattern is easy to see. # The first term is two. # The second term is two times two. # The third term is two times three. # The fourth term is two times four. # The tenth term is two times ten. # the nineteenth term is two times nineteen. # The nth term is two times n. In this sequence the nth term is 2n.

Look up Alfred Wengener, he was the first to descover this!!!!!!!!!!!!!!To prove the idea of Alfred L. Wegener HERE:Wegener used to fit of the continents, the distribution of fossils, the similar sequence of rocks at numerous locations, ancient climates, and the movement of the polar regions.I'm sure with my answer! obejerojamjam@yahoo.com :))

Given two sets of angles and the included side congruent, we seek a sequence of rigid motions that will map Δ_____onto Δ___ proving the triangles congruent.

no prove....