Question

prove that √2 lies between a/b and a+2b/a+b​

Answer 1

Step-by-step explanation:

1<2<9/4 => 1<√2<3/2

Let a<b and a>0 b>0 then a/b <1 so a/b <1<√2……….(1)

√2 < 3/2< (a+2b)/(a+b) [see note below] …….. (2)

from (1) and (2) a/b <1<√2 < 3/2< (a+2b)/(a+b)

note: (a+2b)/(a+b) - 3/2 = [(2a+4b)-(3a+3b)]/2(a+b) = (-a+b)/2(a+b) > 0 as a<b

Answer 2

Answer:

Let us put condition on the second fraction, i.e., a+2ba+b . First, let us assume a+2ba+b>2–√ .

a+2ba+b>2–√,

⟹a+2b>2–√(a+b),

⟹2–√b(2–√−1)>a(2–√−1),

⟹2–√b>a,

⟹ab<2–√.

So, ab<2–√<a+2ba+b .

Similarly by assuming a+2ba+b<2–√ , we will get a+2ba+b<2–√<ab .

So, 2–√ lies either in the interval [ab,a+2ba+b] or [a+2ba+b,ab] for a>0 , b>0 and ab≠2–√.

Step-by-step explanation:

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