{"id":10429,"date":"2022-04-09T20:39:57","date_gmt":"2022-04-09T15:09:57","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10429"},"modified":"2022-04-09T22:32:22","modified_gmt":"2022-04-09T17:02:22","slug":"prove-that-the-following-are-irrationals","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-the-following-are-irrationals\/","title":{"rendered":"Prove that the following are irrationals :"},"content":{"rendered":"

Question<\/strong> : Prove that the following are irrationals :<\/p>\n

(i)  \\(1\\over \\sqrt{2}\\)<\/p>\n

(ii)  \\(7\\sqrt{5}\\)<\/p>\n

(iii)  \\(6 + \\sqrt{2}\\)<\/p>\n

Solution<\/strong> :<\/p>\n

(i) <\/strong> Let us assume, to the contrary, that \\(1\\over \\sqrt{2}\\) is rational. That is, we can find co-prime integers p and q (q \\(\\ne\\) 0) such that<\/p>\n

\\(1\\over \\sqrt{2}\\) = \\(p\\over q\\)  or   \\(1\\times \\sqrt{2}\\over \\sqrt{2}\\times \\sqrt{2}\\) = \\(p\\over q\\) or  \\(\\sqrt{2}\\over 2\\) = \\(p\\over q\\)<\/p>\n

or  \\(\\sqrt{2}\\) = \\(2p\\over q\\)<\/p>\n

Since p and q are integers \\(2p\\over q\\) is rational, and so \\(\\sqrt{2}\\) is rational.<\/p>\n

But this contradicts the fact that \\(\\sqrt{2}\\) is irrational.<\/p>\n

so, we conclude that \\(1\\over \\sqrt{2}\\) is an irrational.<\/strong><\/p>\n

(ii)<\/strong>  Let us assume, to the contrary, that \\(7\\sqrt{5}\\) is rational. That is, we can find co-prime integers p and q (q \\(\\ne\\) 0) such that \\(7\\sqrt{5}\\) = \\(p\\over q\\).<\/p>\n

So,  \\(\\sqrt{5}\\) = \\(p\\over 7q\\)<\/p>\n

Since p and q are integers, \\(p\\over 7q\\) is rational and so is \\(\\sqrt{5}\\).<\/p>\n

But this contradicts the fact that \\(\\sqrt{5}\\) is irrational. So, we conclude that \\(7\\sqrt{5}\\) is an irrational.<\/strong><\/p>\n

(iii)<\/strong>  Let us assume, to the contrary, that \\(6 + \\sqrt{2}\\) is rational. That is, we can find co-prime integers p and q (q \\(\\ne\\) 0) such that<\/p>\n

\\(6 + \\sqrt{2}\\) = \\(p\\over q\\)  or   6 – \\(p\\over q\\) = \\(\\sqrt{2}\\)<\/p>\n

or  \\(\\sqrt{2}\\) = 6 – \\(p\\over q\\)<\/p>\n

Since p and q are integers, 6 – \\(p\\over q\\) is rational, and so \\(\\sqrt{2}\\) is rational.<\/p>\n

But this contradicts the fact that \\(\\sqrt{2}\\) is irrational.<\/p>\n

so, we conclude that \\(6 + \\sqrt{2}\\) is an irrational.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Question : Prove that the following are irrationals : (i)  \\(1\\over \\sqrt{2}\\) (ii)  \\(7\\sqrt{5}\\) (iii)  \\(6 + \\sqrt{2}\\) Solution : (i)  Let us assume, to the contrary, that \\(1\\over \\sqrt{2}\\) is rational. That is, we can find co-prime integers p and q (q \\(\\ne\\) 0) such that \\(1\\over \\sqrt{2}\\) = \\(p\\over q\\)  or   \\(1\\times \\sqrt{2}\\over …<\/p>\n

Prove that the following are irrationals :<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","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 the following are irrationals : - 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-the-following-are-irrationals\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prove that the following are irrationals : - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Question : Prove that the following are irrationals : (i)  (1over sqrt{2}) (ii)  (7sqrt{5}) (iii)  (6 + sqrt{2}) Solution : (i)  Let us assume, to the contrary, that (1over sqrt{2}) is rational. 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