for which pair of integer a b does a+b lie between a and b
Answers
Answer:
we do not know whether
b
a
<
a+b
a+2b
or
b
a
>
a+b
a+2b
Therefore, to compare these two numbers, let us compute
b
a
−
a+b
a+2b
We have,
b
a
−
a+b
a+2b
=
b(a+b)
a(a+b)−b(a+2b)
b(a+b)
a
2
+ab−ab−2b
2
=
b(a+b)
a
2
−2b
2
∴
b
a
−
a+b
a+2b
>0
⇒
b(a+b)
a
2
−2b
2
>0
⇒a
2
−2b
2
>0
⇒a
2
>2b
2
⇒a>
2
b
and,
b
a
−
a+b
a+2b
<0
⇒
b(a+b)
a
2
−2b
2
<0
⇒a
2
−2b
2
<0
⇒a
2
<2b
2
⇒a<
2
b
Thus,
b
a
>
a+b
a+2b
, if a>
2
b and
b
a
<
a+b
a+2b
if a<
2
b.
So, we have the following cases:
Case I when a>
2
b
In this case, we have
b
a
>
a+b
a+2b
i.e.,
a+b
a+2b
<
b
a
We have to prove that
a+b
a+2b
<
2
<
b
a
We have,
a>
2b
⇒a
2
>2b
2
⇒a
2
+a
2
>a
2
+2b
2
[adding a
2
both sides]
⇒2a
2
+2b
2
>(a
2
+2b
2
)+2b
2
[adding 2b
2
on both sides]
⇒2(a
2
+2ab+b
2
)>a
2
+4ab+4b
2
[adding 4ab both sides]
⇒2(a+b)
2
>(a+2b)
2
⇒
2
(a+b)>a+2b
⇒
2
>
a+b
a+2b
Again,
a>
2
b⇒
b
a
>
2
From (i) and (ii), we get
a+b
a+2b
<
2
<
b
a
Case II when a<
2
b
In this case, we have
b
a
<
a+b
a+2b
We have to show that
b
a
<
2
<
a+b
a+2b
We have,
a<
2
b
⇒a
2
<2b
2
⇒a
2
+a
2
<a
2
+2b
2
[adding a
2
on both sides]
⇒2a
2
+2b
2
<a
2
+4b
2
[adding 2b
2
on both sides]
⇒2a
2
+4ab+2b
2
<a
2
+4ab+4b
2
⇒2(a+b)
2
<(a+2b)
2
⇒
2
<
a+b
a+2b
⇒a<
2
b⇒
b
a
<
2
From (iii) and (iv), we get
b
a
<
2
<
a+b
a+2b
Hence
2
lies between
b
a
and
a+b
a+2b