Let a and b be positive integers. Show that √2 always lies between 𝑎𝑏 𝑎𝑛𝑑 𝑎+2𝑏/ 𝑎+𝑏

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We do not know whether ab<a+2ba+b  or ab>a+2ba+b Therefore, to compare these two numbers, let us compute ab-a+2ba+b We have,  ab-a+2ba+b = aa+b-ba+2bba+b = a2+ab-ab-2b2ba+b = a2-2b2ba+b ab-a+2ba+b>0a2-2b2ba+b>0a2-2b>0a2>2ba>2band ab-a+2ba+b<0a2-2b2ba+b<0a2-2b<0a2<2ba<2bThus, ab>a+2ba+b, if a>2b and ab<a+2ba+b, if a<2 bSo, we have the following cases:Case 1: When a>2bIn this cas, we have ab>a+2ba+b i.e., a+2ba+b<abWe have to prove that a+2ba+b<2<abWe have, a>2ba2>2b2a2+a2>a2+2b2                                           adding a2 both sides2a2+2b2>a2+2b2+2b2                         adding 2b2 both sides2a2+2b2+4ab>a2+2b2+2b2+4ab   adding 4ab both sides2a2+b2+2ab>a2+4ab+4b22a+b2>a+2b22a+b>a+2b2>a+2ba+bAgain,a>2bab>2We get,a+2ba+b<2<abCase 2: When a<2bIn this case, we haveab<a+2ba+bWe have to show that ab<2<a+2ba+bWe have,a<2ba2<2b2a2+a2<a2+2b2                                            adding a2 both sides2a2+2b2<a2+2b2+2b2                          adding 2b2 both sides2a2+2b2+4ab<a2+2b2+2b2+4ab     adding 4ab both sides2a2+b2+2ab<a2+4ab+4b22a+b2<a+2b22a+b<a+2b2<a+2ba+bAgain, a<2bab<2We get,ab<2<a+2ba+bHence, 2 lies between ab and a+2ba+b


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Dear student,

If a and b are positive integers. Show that√2 always lies between a by b  and a²-2b² by b(a+b) - Brainly.in

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