{"id":10408,"date":"2022-04-06T22:31:37","date_gmt":"2022-04-06T17:01:37","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10408"},"modified":"2022-04-06T22:31:41","modified_gmt":"2022-04-06T17:01:41","slug":"prove-that-3-2sqrt5-is-irrational","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-3-2sqrt5-is-irrational\/","title":{"rendered":"Prove that \\(3 + 2\\sqrt{5}\\) is irrational."},"content":{"rendered":"

Solution :<\/h2>\n

Let us assume, to the contrary, that \\(3 + 2\\sqrt{5}\\) is an irrational number.<\/p>\n

Now, let \\(3 + 2\\sqrt{5}\\) = \\(a\\over b\\), where a and b are coprime and b \\(ne\\) 0.<\/p>\n

So, \\(2\\sqrt{5}\\) = \\(a\\over b\\) – 3\u00a0 or\u00a0 \\(\\sqrt{5}\\) = \\(a\\over 2b\\) – \\(3\\over 2\\)<\/p>\n

Since a and b are integers, therefore<\/p>\n

\\(a\\over 2b\\) – \\(3\\over 2\\)\u00a0 is a rational number.<\/p>\n

\\(\\therefore\\)\u00a0 \\(\\sqrt{5}\\)\u00a0 is an irrational number.<\/p>\n

But \\(\\sqrt{5}\\)\u00a0 is an irrational number.<\/p>\n

This shows that our assumption is incorrect.<\/p>\n

So, \\(3 + 2\\sqrt{5}\\) is an irrational number.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Let us assume, to the contrary, that \\(3 + 2\\sqrt{5}\\) is an irrational number. Now, let \\(3 + 2\\sqrt{5}\\) = \\(a\\over b\\), where a and b are coprime and b \\(ne\\) 0. So, \\(2\\sqrt{5}\\) = \\(a\\over b\\) – 3\u00a0 or\u00a0 \\(\\sqrt{5}\\) = \\(a\\over 2b\\) – \\(3\\over 2\\) Since a and b are integers, …<\/p>\n

Prove that \\(3 + 2\\sqrt{5}\\) is irrational.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,909],"tags":[],"yoast_head":"\nProve that \\(3 + 2\\sqrt{5}\\) is irrational. - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/prove-that-3-2sqrt5-is-irrational\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prove that \\(3 + 2\\sqrt{5}\\) is irrational. - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Let us assume, to the contrary, that (3 + 2sqrt{5}) is an irrational number. Now, let (3 + 2sqrt{5}) = (aover b), where a and b are coprime and b (ne) 0. So, (2sqrt{5}) = (aover b) – 3\u00a0 or\u00a0 (sqrt{5}) = (aover 2b) – (3over 2) Since a and b are integers, … Prove that (3 + 2sqrt{5}) is irrational. 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