{"id":757,"date":"2017-09-06T22:23:37","date_gmt":"2017-09-07T02:23:37","guid":{"rendered":"http:\/\/blog.richmond.edu\/math320\/?p=757"},"modified":"2017-09-07T08:53:09","modified_gmt":"2017-09-07T12:53:09","slug":"daily-definitions-of-95","status":"publish","type":"post","link":"https:\/\/blog.richmond.edu\/math320\/2017\/09\/06\/daily-definitions-of-95\/","title":{"rendered":"Daily definitions of 9\/5"},"content":{"rendered":"

In Tuesday’s class we covered some challenge questions and some definitions and in this blog I will discuss about the following definitions that we addressed in class: Supremum and maximum, Theorem 1.4.2(Archimedean Property), Theorem 1.4.3(Density of Q in R) and Theorem 1.4.5 (There exists a real number \u03b1 \u2208 R satisfying \u03b1^2 = 2).<\/strong><\/p>\n

1.Supremum and maximum<\/strong><\/p>\n

Supremum<\/strong>: A real number s is the least upper bound for a set A\u00a0\u2286 R if\u00a0it meets the following two criteria:
\n(i) s is an upper bound for A;
\n(ii) if b is any upper bound for A, then s \u2264 b.<\/p>\n

Maximum<\/strong>:\u00a0A real number x is a maximum of the set A if x is an element of A and x \u2265 a for all a \u2208 A.<\/p>\n

Things to consider:<\/em><\/span><\/p>\n

A maximum is a specific type of upper bound that is an element of the set while a supremum is not required to be an element of the set. The open interval (0, 2) does not possess 2 so the open interval (0,2) has a supremum of 2 but it doesn’t have a maximum.\u00a0Thus, the supremum can exist and not be a maximum in this set, but when a maximum exists, then it is also the supremum.<\/p>\n

 <\/p>\n

2. Archimedean Property<\/strong><\/p>\n

(i) Given any number x \u2208 R,
\nthere exists an n \u2208 N satisfying n > x.
\n(ii) Given any real number y > 0, there exists an n \u2208 N satisfying 1\/n < y.<\/p>\n

Pf template:<\/span><\/p>\n

Proving Archimedean Property relies on the Axiom of Completeness(AoC). For part (i) we can assume that N is bounded above and by AoC N should have a least upper bound. So we can set a = supN and a-1 will not be an upper bound since a is the least upper bound for N. So there exists an n\u2208 N satisfying a-1 < n. That means a < n + 1 but n + 1\u00a0\u2208 N so we now have a contradiction that a is a least upper bound.<\/p>\n

And for part (ii) we can set x=1\/y. Since y is a real number and real number is closed under division we know that x is also a real number. Following part(i) we know that for any real number there exists a natural number that is larger than it, which means that 1\/y < n for some n\u00a0\u2208 N. Then we can get 1\/n < y by multiplying both sides by y and 1\/n.<\/p>\n

 <\/p>\n

3. Density of Q in R<\/strong><\/p>\n

For every two real numbers a and b with a < b, there exists a rational number r satisfying a < r < b.<\/p>\n

Pf template:<\/span><\/p>\n

The formula can be changed to a < p\/q < b for some p\u00a0\u2208 Z and q\u00a0\u2208 N. The basic steps of proving this theorem is that first we need to choose a denominator q large enough so that the consecutive increments of size 1\/q will be close together enough to “step over” the interval of (a,b). Using the Archimedean Property to find a p large enough so that 1\/p < b – a. The inequality a < p\/q < b is equivalent to qa < p < qb. Then the idea is to find the smallest integer p so that it will be greater than qa, which means p-1< qa < p. And by transformation and combination of the two inequalities above (…) we should be able to come up with a < p\/q and p\/q < b. Combine these two we get a < p\/q < b as desired.<\/p>\n

 <\/p>\n

Zehao Dong<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"

In Tuesday’s class we covered some challenge questions and some definitions and in this blog I will discuss about the following definitions that we addressed in class: Supremum and maximum, Theorem 1.4.2(Archimedean Property), Theorem 1.4.3(Density of Q in R) and Theorem 1.4.5 (There exists a real number \u03b1 \u2208 R satisfying \u03b1^2 = 2). 1.Supremum […]<\/p>\n","protected":false},"author":3538,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[58823],"tags":[],"class_list":["post-757","post","type-post","status-publish","format-standard","hentry","category-definitions"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7L4E1-cd","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/757","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/users\/3538"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/comments?post=757"}],"version-history":[{"count":0,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/757\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/media?parent=757"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/categories?post=757"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/tags?post=757"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}