[{"id":247536,"date":"2026-08-04T08:18:58","date_gmt":"2026-08-04T13:18:58","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=247536"},"modified":"2026-08-04T08:18:58","modified_gmt":"2026-08-04T13:18:58","slug":"metallic-alchemy","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/08\/04\/metallic-alchemy\/","title":{"rendered":"Metallic alchemy"},"content":{"rendered":"

After writing the previous post<\/a> about metallic ratios, I thought about the analogy to alchemy and the attempt to make precious metals out of base metals.<\/p>\n

When can you make one metallic ratio out of another? Can you make the golden ratio out of the lead ratio?<\/p>\n

Before we can make gold out of lead, we have to say what lead is.<\/p>\n

Defining metallic ratios<\/h2>\n

The metallic ratios\u00a0M<\/em>(n<\/em>) can be defined several ways. The most interesting definition is the number whose continued fraction representation contains all n<\/em>s. A more prosaic but more convenient definition is the larger number that equals its reciprocal plus n<\/em>, which can be found using the quadratic formula.<\/p>\n

\"M(n)<\/p>\n

The golden ratio is\u00a0M<\/em>(1), the silver ratio is\u00a0M<\/em>(2), and the bronze ratio is\u00a0M<\/em>(3).<\/p>\n

Gold from silver and bronze?<\/h2>\n

Can you make the golden ratio out of the silver and bronze ratios? Not by integer arithmetic. The golden ratio involves \u221a5, the silver ratio \u221a2 and the bronze ratio \u221a13. No integer operations on the latter two radicals will produce the former, though you can come arbitrarily close.<\/p>\n

Gold from lead<\/h2>\n

The metallic ratios for n<\/em> > 3 don’t have standard names, but let’s call M<\/em>(4) the lead ratio. Can you make the golden ratio out of the lead ratio? Yes you can:<\/p>\n

M<\/em>(1) = (M<\/em>(4) \u2212 1)\/2.<\/p>\n

General solution<\/h2>\n

In general, when can you make\u00a0M<\/em>(n<\/em>) out of\u00a0M<\/em>(m<\/em>)? In abstract terms the question is when the fields<\/p>\n

\u211a(\u221a(n<\/em>\u00b2 + 4))<\/p>\n

and<\/p>\n

\u211a(\u221a(m<\/em>\u00b2 + 4))<\/p>\n

are the same, i.e. when adjoining \u221a(n<\/em>\u00b2 + 4) to the rational numbers gives the same field as adjoining \u221a(m<\/em>\u00b2 + 4) to the rational numbers. This occurs if and only if<\/p>\n

(n<\/em>\u00b2 + 4)\/(m\u00b2<\/em> + 4)<\/p>\n

is the square of a rational number.<\/p>\n

Bronze from copper and tin<\/h2>\n

Can you make bronze out of copper and tin? Yes, if you define\u00a0M<\/em>(36) to be the copper ratio and\u00a0M<\/em>(393) to be the tin ratio, because<\/p>\n

(3\u00b2 + 4)\/(36\u00b2 + 4) = (1\/10)\u00b2<\/p>\n

and<\/p>\n

(3\u00b2 + 4)\/(292\u00b2 + 4) = (1\/109)\u00b2.<\/p>\n","protected":false},"excerpt":{"rendered":"

After writing the previous post about metallic ratios, I thought about the analogy to alchemy and the attempt to make precious metals out of base metals. When can you make one metallic ratio out of another? Can you make the golden ratio out of the lead ratio? Before we can make gold out of lead, […]<\/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":[94],"class_list":["post-247536","post","type-post","status-publish","format-standard","hentry","category-math","tag-number-theory"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\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\n","aioseo_head_json":{"title":"Metallic alchemy: making one metallic ratio from another","description":"By analogy with alchemy, can you make gold from lead? i.e. can you make the golden ratio by integer operations on the lead ratio?","canonical_url":"https:\/\/www.johndcook.com\/blog\/2026\/08\/04\/metallic-alchemy\/","robots":"max-image-preview:large","keywords":"number theory","webmasterTools":{"miscellaneous":""},"schema":null,"og:locale":"en_US","og:site_name":"John D. Cook | Applied Mathematics Consulting","og:type":"article","og:title":"Metallic alchemy: making one metallic ratio from another","og:description":"By analogy with alchemy, can you make gold from lead? i.e. can you make the golden ratio by integer operations on the lead ratio?","og:url":"https:\/\/www.johndcook.com\/blog\/2026\/08\/04\/metallic-alchemy\/","article:published_time":"2026-08-04T13:18:58+00:00","article:modified_time":"2026-08-04T13:18:58+00:00","twitter:card":"summary","twitter:title":"Metallic alchemy: making one metallic ratio from another","twitter:description":"By analogy with alchemy, can you make gold from lead? i.e. can you make the golden ratio by integer operations on the lead ratio?","twitter:image":"https:\/\/www.johndcook.com\/blog\/wp-content\/uploads\/2022\/05\/twittercard.png"},"aioseo_meta_data":{"post_id":"247536","title":"Metallic alchemy: making one metallic ratio from another","description":"By analogy with alchemy, can you make gold from lead? i.e. can you make the golden ratio by integer operations on the lead ratio?","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":"2026-08-04 12:23:52","updated":"2026-08-04 18:09:02","ai":{"faqs":[],"keyPoints":[],"schemas":[],"titles":[],"descriptions":[],"socialPosts":{"email":{"subject":"","preview":"","content":""},"linkedin":[],"twitter":[],"facebook":[],"instagram":[]}},"seo_analyzer_scan_date":null},"aioseo_breadcrumb":"

\n\t\t\tHome<\/a>\n\t\t<\/span>»<\/span>\n\t\t\tMath<\/a>\n\t\t<\/span>»<\/span>\n\t\t\tMetallic alchemy\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":"Metallic alchemy","link":"https:\/\/www.johndcook.com\/blog\/2026\/08\/04\/metallic-alchemy\/"}],"_links":{"self":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247536","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=247536"}],"version-history":[{"count":2,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247536\/revisions"}],"predecessor-version":[{"id":247538,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247536\/revisions\/247538"}],"wp:attachment":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/media?parent=247536"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/categories?post=247536"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/tags?post=247536"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}},{"id":247532,"date":"2026-08-04T06:41:49","date_gmt":"2026-08-04T11:41:49","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=247532"},"modified":"2026-08-04T06:41:49","modified_gmt":"2026-08-04T11:41:49","slug":"ratio-of-metallic-ratios","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/08\/04\/ratio-of-metallic-ratios\/","title":{"rendered":"Ratio of metallic ratios"},"content":{"rendered":"

The golden ratio is the first and best known of the metallic ratios. I’ve written about the silver ratio a few times, most recently here<\/a>. And I’ve mentioned the bronze ratio<\/a> a couple times. The metallic ratios after bronze don’t have standard names.<\/p>\n

The n<\/em>th metallic ratio M<\/em>(n<\/em>) is the number whose continued fraction representation contains all n<\/em>s.<\/p>\n

\"n<\/p>\n

When n<\/em> = 1, 2, and 3 we get the gold, silver, and bronze ratios.<\/p>\n

You can approximate any positive real number as a ratio of metallic ratios. To see this, note that for large\u00a0n<\/em>, M<\/i>(n<\/em>)\u00a0is approximately\u00a0n<\/em>. For any positive rational number a<\/em>\/b<\/em>,<\/p>\n

\"\\lim_{n\\to\\infty}<\/p>\n

and so you can make\u00a0M<\/em>(na<\/em>) \/\u00a0M<\/em>(nb<\/em>) as close to\u00a0a<\/em>\/b<\/em> as you like by taking\u00a0n<\/em> large enough. And since the rationals are dense in the reals, you can approximate any positive real number as close as you’d like.<\/p>\n

Let’s look for metallic ratios whose ratios approximate \u03c0 to within 0.001 with the following Python code.<\/p>\n

from math import pi, sqrt\r\n\r\nM = lambda n: 0.5*(n + sqrt(n**2 + 4))\r\n\r\nfor n in range(1, 100):\r\n    a = round(pi*n)\r\n    b = n\r\n    r = M(a)\/M(b)\r\n    if abs(r - pi) < 0.001:\r\n        print(a, b, r)\r\n<\/pre>\n
This shows<\/p>\n
\u03c0 \u2248\u00a0M<\/em>(132) \/\u00a0M<\/em>(42) = 3.1412\u2026<\/p>\n
Could we find smaller numbers that work? The following code shows the answer is no.<\/p>\n
k = 132 + 42\r\n# loop over numbers whose sum is less than k\r\nfor n in range(1, k):\r\n    for a in range(1, n):\r\n        b = n - a\r\n        r = M(a)\/M(b)\r\n        if abs(r - pi) < 0.001:\r\n            print(a, b, r)\r\n            exit()\r\n<\/pre>\n
Related posts<\/h2>\n