If you’re interested in some math recreation: the square root of 2 is a number like pi. It’s irrational, meaning that its decimal expansion does not terminate or repeat. It goes on forever with no pattern. You could never fully specify it as a decimal. Numbers that do have a terminal or repeating decimal expansion are called rational numbers. <\/p>\n
Proof<\/strong> (I will be longwinded): <\/p>\n Claim<\/strong>: if a number has a terminal or repeating decimal expansion, then it can be written as A\/B, where A and B are whole numbers. <\/p>\n Non-rigorous Proof of Claim<\/strong>: The case of the terminal expansion is easy. A decimal that ends, such as 0.12345 is just 12345\/100,000. <\/p>\n In the case of the repeating decimal, the proof involves a trick. Suppose a number S = 0.abcabcabcabc… repeating abc forever. <\/p>\n Then, 1000*S = abc.abcabcabc… repeating abc forever. <\/p>\n 1000*S – S = abc.abcabc… – 0.abcabcabc… = abc (if you don’t believe it, let a = 1, b = 2, c = 3)<\/p>\n Therefore, 1000*S – S = abc. <\/p>\n 1000*S – S = 999*S (if S is an apple, this is like saying 1000 apples – 1 apple is 999 apples). <\/p>\n 999*S = abc<\/p>\n This means that S = abc\/999 and we have converted the repeating decimal S into a fraction of the form A\/B. A similar trick works for any repeating decimal. <\/p>\n Exercise<\/strong>: try proving that the repeating decimal 0.1111… is a simple fraction. What is it?<\/p>\n The proof that the square root of 2 is irrational hinges on showing that it is impossible to write the square root of 2 as a fraction A\/B. In other words, the square root of 2 cannot rational. This is a technique called proof by contradiction. It’s often used in squirrelly matters such as infinity. <\/p>\n Proofs by contradiction all have a similar format: suppose something outrageous. Prove that it violates logic. If every step of the argument is logically sound, then the original outrageous assumption must be the one that is false. <\/p>\n Outrageous assumption: the square root of 2 has a terminating or repeating decimal. <\/p>\n Implication: if true, then there are numbers A and B where the square root of 2 is A\/B. <\/p>\n Implication: furthermore, it’s possible to find A and B where the fraction A\/B is in lowest terms, in other words, A and B have no common factors. (You know from elementary school math that common factors can be canceled out. So we assume they are canceled out.)<\/p>\n (refresher: 1\/3 is in lowest terms. 2\/6 is not in lowest terms). <\/p>\n If A\/B = square root of 2, <\/p>\n then (A\/B) * (A\/B) = 2<\/p>\n This means that A*A \/ B*B = 2<\/p>\n This means that A*A = 2*B*B. Claim<\/strong>: The only way this can happen is if 2 is a factor of A. The number A must be of the form A=2*K. <\/p>\n That means that A*A = (2K)*(2K) = 4 K*K. <\/p>\n If 4*K*K = 2 B*B, <\/p>\n then 2*K*K = B*B. <\/p>\n The only way this can happen is if 2 is a factor of B. <\/p>\n However, this violates the implication that A and B have no common factors. (This is a more detailed version of the proof you find on wikipedia.)<\/p>\n","protected":false},"excerpt":{"rendered":" If you’re interested in some math recreation: the square root of 2 is a number like pi. It’s irrational, meaning that its decimal expansion does not terminate or repeat. It goes on forever with no pattern. You could never fully specify it as a decimal. Numbers that do have a terminal or repeating decimal expansion […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-57","post","type-post","status-publish","format-standard","hentry","category-proof"],"_links":{"self":[{"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/posts\/57","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=57"}],"version-history":[{"count":10,"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/posts\/57\/revisions"}],"predecessor-version":[{"id":77,"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=\/wp\/v2\/posts\/57\/revisions\/77"}],"wp:attachment":[{"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=57"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=57"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bear-mathbelt.marjoriesayer.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=57"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nThis in turn means that 2 is a factor of A*A. <\/p>\n
\nProof of Claim<\/strong>: Let S*T be the product of two numbers. Then either:
\nS is even, which makes S*T even; or,
\nT is even, making S*T even; or,
\nS and T are both even, making S*T even, or
\nS and T are both odd, making S*T odd.
\nIn the special case S = T = A, and A*A is even, then A must be even, so 2 is a factor of A. <\/p>\n
\nNumbers A and B that satisfy the original assumption cannot be found. <\/p>\n