If a and b are positive integers. Show that √2 always lies between a and a − 2b b a + b a2 −2b2 a a + 2b

Ans:  We do not know whether   or   <  

b(a+b) b a + b
   ∴ to compare these two number,

    Let us comute  a   − a + 2b 

b a + b
a2 −2b2
        => on simplifying , we get  

b(a+b)

 ∴ a   − a + 2b > 0 or a − a + 2b < 0 b a + b b a + b



  now a   − a + 2b   > 0 b a + b


  a2 −2b2  > 0   solve it , we get ,  a > √2b b(a+b)


Thus  , when a > √2b and  a a + 2b
< ,  b a + b


a + 2b   √2< a
We have to prove that  <

a + b b

Now a    >√2 b⇒2a2+2b2>2b2+ a2+2b2
On simplifying   we get 

        √2> a + 2b a + b

Also a>√2
>√2  b
Similarly we get √2, < a + 2b
a + b


Hence a   <√2< a + 2b b a + b