The Real And Complex Number Systems

In addition,
(√ a
)
2 − −
b
( a + 2b
a + b − √ )
2 = 2 √ 2 −
( a
b + a + 2b )
a + b
= 2 √ 2 − a2 + 2ab + 2b 2
ab + b 2
1
[(
= 2 √ )
2 − 2 ab +
ab + b 2
≥
= 0.
1
ab + b 2 [ (
2 √ 2 − 2
)
a√ a + 2
(
2 √ ) ]
2 − 2 b 2 − a 2
(
2 √ ) ( ) ] 2
a
2 − 2 √2 − a 2
So, a+2b is closer to √ 2.
a+b
Similarly, we also have if a/b > √ 2, then a+2b
a+b
to √ 2 in this case.
Remark: Note that
a
b < √ 2 < a + 2b
a + b < 2b
a
< √ 2. Also, a+2b
a+b
by Exercise 12 and 13.
is closer
And we know that a+2b is closer to √ 2. We can use it to approximate √ 2.
a+b
Similarly for the case
2b
a < a + 2b
a + b < √ 2 < a b .
1.14 Prove that √ n − 1 + √ n + 1 is irrational for every integer n ≥ 1.
Proof: Suppose that √ n − 1 + √ n + 1 is rational, and thus consider
( √n
+ 1 +
√
n − 1
) ( √n
+ 1 −
√
n − 1
)
= 2
which implies that √ n + 1 − √ n − 1 is rational. Hence, √ n + 1 and √ n − 1
are rational. So, n − 1 = k 2 and n + 1 = h 2 , where k and h are positive
integer. It implies that
h = 3 2 and k = 1 2
which is absurb. So, √ n − 1 + √ n + 1 is irrational for every integer n ≥ 1.
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