[{"id":248181,"date":"2026-11-03T08:41:25","date_gmt":"2026-11-03T14:41:25","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=248181"},"modified":"2026-11-03T08:41:25","modified_gmt":"2026-11-03T14:41:25","slug":"fraction-comparison-trick-revisited","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/11\/03\/fraction-comparison-trick-revisited\/","title":{"rendered":"Fraction comparison trick revisited"},"content":{"rendered":"

A few years ago<\/a> I wrote about a curious result: a<\/em>\/b<\/em>\u00a0is usually greater than\u00a0c<\/em>\/d<\/em>\u00a0when\u00a0a\u00a0<\/em>+\u00a0d<\/em>\u00a0is greater than\u00a0b\u00a0<\/em>+\u00a0c<\/em>. And “usually” means with probability 11\/12.<\/p>\n

As stated in the post, you can make this statement rigorous by choosing integers up to size\u00a0N<\/em> at random and taking the limit as\u00a0N<\/em> goes to infinity.<\/p>\n

We can also choose\u00a0a<\/em>,\u00a0b<\/em>,\u00a0c<\/em>, and\u00a0d<\/em> uniformly from [0, 1]. This is similar to choosing four integers up to N<\/em> and dividing all the integers by\u00a0N<\/em>, which doesn’t change the probabilities.<\/p>\n

Now\u00a0a<\/em>\/b<\/em> >\u00a0c<\/em>\/d<\/em> if and only if\u00a0ad<\/em> >\u00a0bc<\/em>, so another way to state the theorem is that<\/p>\n

Prob(ad<\/em> >\u00a0bc <\/em>| a<\/em> +\u00a0d<\/em> >\u00a0b<\/em> +\u00a0c<\/em>) = 11\/12.<\/p>\n

Or to state it yet another way, pick two pairs of numbers at random. The probability that the pair with the larger sum also has the larger product is 11\/12. That seems more intuitive to me than the original statement.<\/p>\n

If a<\/em> +\u00a0d<\/em> is greater than\u00a0b<\/em> +\u00a0c<\/em>, it’s likely that\u00a0a<\/em>\/b<\/em> is greater than c<\/em>\/d<\/em>. It seems plausible that if\u00a0a<\/em> +\u00a0d<\/em> is much<\/em> greater than b<\/em> +\u00a0c<\/em>, it’s more<\/em> likely that a<\/em>\/b<\/em> is greater than c<\/em>\/d<\/em>. Or in terms of cross products, the more the sum of one pair is bigger than the sum of the other, the more likely the product of the former is to be greater than the latter. That is<\/p>\n

f<\/em>(r<\/em>) = Prob(ad<\/em> >\u00a0bc <\/em>| (a<\/em> +\u00a0d<\/em>)\/(b<\/em> +\u00a0c<\/em>) \u2265 r<\/em>)<\/p>\n

should be an increasing function of\u00a0r<\/em>. We know the probability above equals 11\/12 when\u00a0r<\/em> = 1. We expect that the probability should approach 1 as r<\/em> goes to infinity.<\/p>\n

This is indeed the case, as shown in the plot below.<\/p>\n

<\/p>\n

In particular, f<\/em>(2) \u2248 0.99 and\u00a0f<\/em>(6) \u2248 0.999. So if\u00a0a<\/em> +\u00a0d<\/em> is two (or six) times bigger than\u00a0b +<\/em>\u00a0c, there is a 0.99 (or 0.999) probability that\u00a0ad<\/em> is larger than\u00a0bc<\/em>.<\/p>\n

You can calculate\u00a0f<\/em>(r<\/em>) analytically or estimate it by simulation. I had Claude do both and they produce comparable results. There is a closed-form expression for\u00a0f<\/em>(r<\/em>) but it’s really complicated.<\/p>\n","protected":false},"excerpt":{"rendered":"

A few years ago I wrote about a curious result: a\/b\u00a0is usually greater than\u00a0c\/d\u00a0when\u00a0a\u00a0+\u00a0d\u00a0is greater than\u00a0b\u00a0+\u00a0c. And “usually” means with probability 11\/12. As stated in the post, you can make this statement rigorous by choosing integers up to size\u00a0N at random and taking the limit as\u00a0N goes to infinity. We can also choose\u00a0a,\u00a0b,\u00a0c, and\u00a0d uniformly […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[9],"tags":[105],"class_list":["post-248181","post","type-post","status-publish","format-standard","hentry","category-math","tag-probability"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\n","aioseo_head_json":{"title":"Fraction comparison trick revisited","description":"Revisiting the theorem that a\/b is usually greater than c\/d when a + d is greater than b + c. Different formulations and an extension.","canonical_url":"https:\/\/www.johndcook.com\/blog\/2026\/11\/03\/fraction-comparison-trick-revisited\/","robots":"max-image-preview:large","keywords":"probability","webmasterTools":{"miscellaneous":""},"schema":null,"og:locale":"en_US","og:site_name":"John D. Cook | Applied Mathematics Consulting","og:type":"article","og:title":"Fraction comparison trick revisited","og:description":"Revisiting the theorem that a\/b is usually greater than c\/d when a + d is greater than b + c. Different formulations and an extension.","og:url":"https:\/\/www.johndcook.com\/blog\/2026\/11\/03\/fraction-comparison-trick-revisited\/","article:published_time":"2026-11-03T14:41:25+00:00","article:modified_time":"2026-11-03T14:41:25+00:00","twitter:card":"summary","twitter:title":"Fraction comparison trick revisited","twitter:description":"Revisiting the theorem that a\/b is usually greater than c\/d when a + d is greater than b + c. Different formulations and an extension.","twitter:image":"https:\/\/www.johndcook.com\/blog\/wp-content\/uploads\/2022\/05\/twittercard.png"},"aioseo_meta_data":{"post_id":"248181","title":null,"description":"Revisiting the theorem that a\/b is usually greater than c\/d when a + d is greater than b + c. Different formulations and an extension.","keywords":null,"keyphrases":{"focus":{"keyphrase":"","score":0,"analysis":{"keyphraseInTitle":{"score":0,"maxScore":9,"error":1}}},"additional":[]},"primary_term":null,"canonical_url":null,"og_title":null,"og_description":null,"og_object_type":"default","og_image_type":"default","og_image_url":null,"og_image_width":null,"og_image_height":null,"og_image_custom_url":null,"og_image_custom_fields":null,"og_video":"","og_custom_url":null,"og_article_section":null,"og_article_tags":null,"twitter_use_og":false,"twitter_card":"default","twitter_image_type":"default","twitter_image_url":null,"twitter_image_custom_url":null,"twitter_image_custom_fields":null,"twitter_title":null,"twitter_description":null,"schema":{"blockGraphs":[],"customGraphs":[],"default":{"data":{"Article":[],"Course":[],"Dataset":[],"FAQPage":[],"Movie":[],"Person":[],"Product":[],"ProductReview":[],"Car":[],"Recipe":[],"Service":[],"SoftwareApplication":[],"WebPage":[]},"graphName":"Article","isEnabled":true},"graphs":[]},"schema_type":"default","schema_type_options":null,"pillar_content":false,"robots_default":true,"robots_noindex":false,"robots_noarchive":false,"robots_nosnippet":false,"robots_nofollow":false,"robots_noimageindex":false,"robots_noodp":false,"robots_notranslate":false,"robots_max_snippet":"-1","robots_max_videopreview":"-1","robots_max_imagepreview":"large","priority":null,"frequency":"default","location":null,"local_seo":null,"breadcrumb_settings":null,"limit_modified_date":false,"created":"2026-11-03 13:45:03","updated":"2026-11-03 18:58:01","ai":{"faqs":[],"keyPoints":[],"schemas":[],"titles":[],"descriptions":[],"socialPosts":{"email":{"subject":"","preview":"","content":""},"linkedin":[],"twitter":[],"facebook":[],"instagram":[]}},"seo_analyzer_scan_date":null,"focus_keyword":null,"additional_keywords":null,"truseo_locale":null},"aioseo_breadcrumb":"

\n\t\t\tHome<\/a>\n\t\t<\/span>\u00bb<\/span>\n\t\t\tMath<\/a>\n\t\t<\/span>\u00bb<\/span>\n\t\t\tFraction comparison trick revisited\n\t\t<\/span><\/div>","aioseo_breadcrumb_json":[{"label":"Home","link":"https:\/\/www.johndcook.com\/blog\/"},{"label":"Math","link":"https:\/\/www.johndcook.com\/blog\/category\/math\/"},{"label":"Fraction comparison trick revisited","link":"https:\/\/www.johndcook.com\/blog\/2026\/11\/03\/fraction-comparison-trick-revisited\/"}],"_links":{"self":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/248181","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/comments?post=248181"}],"version-history":[{"count":1,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/248181\/revisions"}],"predecessor-version":[{"id":248182,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/248181\/revisions\/248182"}],"wp:attachment":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/media?parent=248181"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/categories?post=248181"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/tags?post=248181"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}},{"id":248173,"date":"2026-11-03T06:33:10","date_gmt":"2026-11-03T12:33:10","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=248173"},"modified":"2026-11-03T06:33:10","modified_gmt":"2026-11-03T12:33:10","slug":"spherical-and-projective-geometry","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/11\/03\/spherical-and-projective-geometry\/","title":{"rendered":"Euclidean, Spherical, and Projective Geometry"},"content":{"rendered":"

Spherical geometry is very tangible because we live on a sphere. Projective geometry seems much more exotic, and yet they’re closely related. To show how they’re connected, let’s go back to the origins of non-Euclidean geometry.<\/p>\n

Euclidean geometry<\/h2>\n

Here are Euclid’s five postulates, statements about what geometric objects can be constructred.<\/p>\n

    \n
  1. To draw a\u00a0straight line\u00a0from any\u00a0point\u00a0to any point.<\/li>\n
  2. To produce (extend) a\u00a0finite straight line\u00a0continuously in a straight line.<\/li>\n
  3. To describe a\u00a0circle\u00a0with any centre and distance (radius).<\/li>\n
  4. That all\u00a0right angles\u00a0are equal to one another.<\/li>\n
  5. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.<\/li>\n<\/ol>\n

    Clearly there’s something different about the fifth postulate, the parallel postulate. It’s much longer and more complicated than the other postulates. There was speculation from ancient times that the parallel postulate might be unnecessary, that it could be derived as a theorem rather than stated as an axiom.<\/p>\n

    Non-Euclidean geometry<\/h2>\n

    Giovanni Girolamo Saccheri (1667\u20131733) tried to prove the fifth postulate by contradiction, and in the process began the study of non-Euclidean geometry. He wasn’t the first to try to prove the fifth postulate. Ptolemy, for example, tried to prove the fifth postulate as a theorem in the second century AD. But Saccheri was the first to try a proof by contradiction, or at least the first to extensively explore contradictions to the parallel postulate.<\/p>\n

    A more modern formulation of the parallel postulate says<\/p>\n

    Given a line \u2113<\/em> and a point P<\/em> not on \u2113<\/em>, there is exactly one line through P<\/em> parallel to \u2113<\/em>.<\/p>\n

    Non-Euclidean geometry replaces this axiom with an alternative, such as saying there are multiple lines parallel to \u2113 <\/em>or that there are no lines parallel to \u2113<\/em>.<\/p>\n

    Note that non-Euclidean geometry is not an arbitrary tinkering with the axioms of geometry just to see what happens. It tinkers with one specific axiom for motivated reasons.<\/p>\n

    Spherical geometry<\/h2>\n

    You can create models of non-Euclidean geometry using Euclidean geometry. For example, spherical geometry is constructed by defining a (spherical) point as a (Euclidean) point on a sphere, and defining a (spherical) line to be a great circle, the intersection of a (Euclidean) plane with the sphere. In spherical geometry there are no parallel lines: all lines intersect.<\/p>\n

    This example shows that non-Euclidean geometry need not be non-physical. If you leave general relativity out of the picture and think of physical space as undeniably Euclidean, it can still be useful to think of different geometries to describe objects in Euclidean space.<\/p>\n

    Projective geometry<\/h2>\n

    In projective geometry, all distinct lines intersect at one point.<\/p>\n

    In spherical geometry, all distinct lines intersect at two points, on opposite sides of a great circle. If you consider antipodal pairs of points on a sphere to be a single point in the projective plane, now all lines intersect at a single point. Projective geometry is just one step away from spherical geometry.<\/p>\n

    The recent thread of posts started with a post<\/a> a couple weeks ago about two versions of the law of cosines on a sphere, one for sides and one for angles. You can go from one to the other by going from a spherical triangle to its polar triangle. That started me going down the rabbit hole of poles and polars, first in spherical geometry, then in plane geometry, then in projective geometry, culminating here<\/a>.<\/p>\n

    Now we have two ways of producing poles and polars on a sphere: the original construction in terms of spherical geometry, and the projective construction. They are consistent: you can do spherical geometry, then identify antipodal points, or you can identify antipodal points, then do projective geometry. You end up at the same place by either path.<\/p>\n

    One theme in the last few posts has been that projective geometry is not an arbitrary exercise in abstraction but a framework that can simplify how we think about things. For example, when I defined the polar triangle of a spherical triangle here<\/a>, I said that the vertices of the polar triangle are the poles of the sides of the original triangle. Imagine the side as the equator of the sphere and take one of the poles; that’s where the abstract concept of “pole” comes from. But which pole do you take? There is a convention for answering this question, but in projective geometry it doesn’t matter: both (spherical) poles constitute one (projective) point.<\/p>\n

    Posts in this series<\/h2>\n