{"id":353,"date":"2019-06-26T23:43:26","date_gmt":"2019-06-26T23:43:26","guid":{"rendered":"http:\/\/funfacts.104.42.120.246.xip.io\/?page_id=353"},"modified":"2019-11-18T23:59:03","modified_gmt":"2019-11-18T23:59:03","slug":"e-is-irrational","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/funfacts\/e-is-irrational\/","title":{"rendered":"e is irrational"},"content":{"rendered":"\n

If e<\/em> were rational, then e<\/em> = n\/m for some integers m, n. So then 1\/e<\/em> = m\/n. But the series expansion for 1\/e<\/em> is<\/p>\n\n\n\n

1\/e<\/em> = 1 – 1\/1! + 1\/2! – 1\/3! + …<\/p>\n\n\n\n

Call the sum of the first n terms of this alternating series S(n). How good is this approximation to 1\/e<\/em>? Well, the error is bounded by the next term of the alternating series:<\/p>\n\n\n\n

0 < | 1\/e<\/em> – S(n) | = | m\/n – S(n)| < 1\/(n+1)!<\/p>\n\n\n\n

But multiplying through by n!, you will see that<\/p>\n\n\n\n

0 < | integer – integer | < 1\/(n+1) < 1.<\/p>\n\n\n\n

But there is no integer strictly between 0 and 1, so this is a contradiction; e must be irrational.<\/p>\n\n\n\n

Presentation\u00a0Suggestions:<\/strong>
Use the\u00a0series\u00a0expansion for 1\/e<\/em>\u00a0as a fun fact on a previous day.<\/p>\n\n\n\n

The\u00a0Math\u00a0Behind\u00a0the\u00a0Fact:<\/strong>
Anytime you have an\u00a0alternating series\u00a0in which the terms decrease, then each\u00a0partial sum\u00a0is not farther from the limit than the next term in the series!<\/p>\n\n\n\n

On an unrelated note, it is much harder to show that Pi is irrational.<\/p>\n\n\n\n

How to Cite this Page:<\/strong> 
Su, Francis E., et al. “e<\/em> is irrational.” Math Fun Facts<\/em>. <http:\/\/www.math.hmc.edu\/funfacts>.<\/p>\n\n\n\n

References<\/strong>:
For more about e, see Maor, e: the story of a number<\/a>.<\/p>\n\n\n\n

Fun Fact suggested by:<\/strong>
Arthur Benjamin<\/p>\n","protected":false},"excerpt":{"rendered":"

If e were rational, then e = n\/m for some integers m, n. So then 1\/e = m\/n. But the series expansion for 1\/e is 1\/e =...<\/p>\n","protected":false},"author":7,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[94,20,7,4,10,51],"class_list":["post-353","page","type-page","status-publish","hentry","tag-alternating-series","tag-analysis","tag-calculus","tag-medium","tag-numtheory","tag-proof-by-contradiction"],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/353","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/comments?post=353"}],"version-history":[{"count":3,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/353\/revisions"}],"predecessor-version":[{"id":1414,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/353\/revisions\/1414"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/media?parent=353"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/tags?post=353"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}