If a and b are two integers, show that root2 lies between a/b and a+2b/a+b
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We do not know whether\xa0{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b } \\text { or, } \\frac { a } { b } > \\frac { a + 2 b } { a + b }{/tex}.Therefore, to compare these two numbers, let us compute\xa0{tex} \\frac { a } { b } – \\frac { a + 2 b } { a + b }{/tex}We have,{tex} \\frac { a } { b } – \\frac { a + 2 b } { a + b } = \\frac { a ( a + b ) – b ( a + 2 b ) } { b ( a + b ) }{/tex}\xa0{tex} = \\frac { a ^ { 2 } + a b – a b – 2 b ^ { 2 } } { b ( a + b ) } = \\frac { a ^ { 2 } – 2 b ^ { 2 } } { b ( a + b ) }{/tex}{tex} \\therefore \\quad \\frac { a } { b } – \\frac { a + 2 b } { a + b } > 0{/tex}{tex} \\Rightarrow \\quad \\frac { a ^ { 2 } – 2 b ^ { 2 } } { b ( a + b ) } > 0{/tex}{tex} \\Rightarrow{/tex} a2 – 2b2 > 0{tex} \\Rightarrow{/tex} a2> 2b2{tex} \\Rightarrow \\quad a > \\sqrt { 2 } b{/tex}and,\xa0{tex} \\frac { a } { b } – \\frac { a + 2 b } { a + b } < 0{/tex}{tex} \\Rightarrow \\quad \\frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) } < 0{/tex}{tex} \\Rightarrow{/tex} a2 - 2b2 < 0{tex} \\Rightarrow{/tex}a2 <2b2{tex} \\Rightarrow \\quad a < \\sqrt { 2 } b{/tex}Thus,\xa0{tex} \\frac { a } { b } > \\frac { a + 2 b } { a + b }{/tex}, if\xa0{tex}a > \\sqrt { 2 b }{/tex}\xa0and\xa0{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b }{/tex},\xa0if\xa0{tex} a < \\sqrt { 2 } b{/tex}.So, we have the following cases:CASE I When\xa0{tex} a > \\sqrt { 2 } b{/tex}In this case, we have{tex} \\frac { a } { b } > \\frac { a + 2 b } { a + b } \\text { i.e., } \\frac { a + 2 b } { a + b } < \\frac { a } { b }{/tex}We have to prove that{tex} \\frac { a + 2 b } { a + b } < \\sqrt { 2 } < \\frac { a } { b }{/tex}We have,{tex} a > \\sqrt { 2 } b{/tex}{tex} \\Rightarrow{/tex} a2> 2b2 [Adding a2 on both sides]{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } > \\left( a ^ { 2 } + 2 b ^ { 2 } \\right) + 2 b ^ { 2 }{/tex}\xa0[Adding\xa02b2 on both sides]{tex} \\Rightarrow \\quad 2 \\left( a ^ { 2 } + b ^ { 2 } \\right) + 4 a b > a ^ { 2 } + 4 b ^ { 2 } + 4 a b{/tex}\xa0[Adding 4ab on both sides]{tex} \\Rightarrow \\quad 2 \\left( a ^ { 2 } + 2 a b + b ^ { 2 } \\right) > a ^ { 2 } + 4 a b + 4 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad 2 ( a + b ) ^ { 2 } > ( a + 2 b ) ^ { 2 }{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } ( a + b ) > a + 2 b{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } > \\frac { a + 2 b } { a + b }{/tex} ……..(i)Again,{tex} a > \\sqrt { 2 } b {/tex}{tex}\\Rightarrow \\frac { a } { b } > \\sqrt { 2 }{/tex} …….(ii)From (i) and (ii), we get{tex} \\frac { a + 2 b } { a + b } < \\sqrt { 2 } < \\frac { a } { b }{/tex}CASE II When\xa0{tex} a < \\sqrt { 2 } b{/tex}In this case, we have{tex} \\frac { a } { b } < \\frac { a + 2 b } { a + b }{/tex}We have to show that\xa0{tex} \\frac { a } { b } < \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}We have,{tex} a < \\sqrt { 2 } b{/tex}{tex} \\Rightarrow \\quad a ^ { 2 } < 2 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad a ^ { 2 } + a ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }{/tex}\xa0[Adding a2 on both sides]{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }+ 2 b ^ { 2 }{/tex}\xa0[Adding 2b2 on both sides]{tex}\\Rightarrow \\quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 4 b ^ { 2 }{/tex}{tex} \\Rightarrow \\quad 2 a ^ { 2 } + 4 a b + 2 b ^ { 2 } < a ^ { 2 } + 4 a b + 4 b ^ { 2 }{/tex}\xa0[Adding 4ab on both sides]{tex} \\Rightarrow \\quad 2 ( a + b ) ^ { 2 } < ( a + 2 b ) ^ { 2 }{/tex}{tex} \\Rightarrow \\sqrt { 2 } ( a + b ) < a + 2 b{/tex}{tex} \\Rightarrow \\quad \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}\xa0. ...(iii){tex} \\Rightarrow \\quad a < \\sqrt { 2 } b \\Rightarrow \\frac { a } { b } < \\sqrt { 2 }{/tex} ....(iv)From (iii) and (iv), we get{tex} \\frac { a } { b } < \\sqrt { 2 } < \\frac { a + 2 b } { a + b }{/tex}Hence,\xa0{tex} \\sqrt { 2 }{/tex}\xa0lies between\xa0{tex} \\frac { a } { b }{/tex}\xa0and\xa0{tex} \\frac { a + 2 b } { a + b }{/tex}.