We do not know whether {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 {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} {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, {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, {tex} \frac { a } { b } > \frac { a + 2 b } { a + b }{/tex}, if {tex}a > \sqrt { 2 b }{/tex} and {tex} \frac { a } { b } < \frac { a + 2 b } { a + b }{/tex}, if {tex} a < \sqrt { 2 } b{/tex}.
So, we have the following cases:
CASE I When {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} [Adding 2b² 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} [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 {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 {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} [Adding a² on both sides]
{tex} \Rightarrow \quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }+ 2 b ^ { 2 }{/tex} [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} [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} . ...(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, {tex} \sqrt { 2 }{/tex} lies between {tex} \frac { a } { b }{/tex} and {tex} \frac { a + 2 b } { a + b }{/tex}.

Given integers are 408 and 1032 where 408 < 1032
By applying Euclid’s division lemma, we get 1032 = 408 {tex}\times{/tex} 2 + 216.
Since the remainder ≠ 0, so apply division lemma again on divisor 408 and remainder 216, we get the relation as
408 = 216 {tex}\times{/tex} 1 + 192.
Since the remainder ≠ 0, so apply division lemma again on divisor 216 and remainder 192
216 = 192 {tex}\times{/tex} 1 + 24.
Since the remainder ≠ 0, so apply division lemma again on divisor 192 and remainder 24 
192 = 24 × 8 + 0.
Now the  remainder has become 0. Therefore, the H.C.F of 408 and 1032 = 24.
Therefore,
24 = 1032m - 408 {tex}\times{/tex} 5
1032m = 24 + 408 {tex}\times{/tex} 5
1032m = 24 + 2040
1032m = 2064  
{tex}m = \frac{{2064}}{{1032}}{/tex}
Therefore, m = 2.

On dividing n by 3, let q be the quotient and r be the remainder.
Then, {tex}n = 3q + r{/tex}, where {tex}0 \leq r < 3{/tex}
{tex}\Rightarrow\;n = 3q + r{/tex} , where r = 0,1 or 2
{tex}\Rightarrow{/tex} {tex}n = 3q \;or \;n = (3q + 1) \;or\; n = (3q + 2){/tex}.
Case I If n = 3q then n is clearly divisible by 3.
Case II If {tex}\;n = (3q + 1)\; {/tex} then {tex} (n + 2)= (3q + 1 + 2) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.
In this case, {tex}(n + 2){/tex} is divisible by 3.
Case III If n = {tex}(3q + 2){/tex} then {tex}(n + 1) = (3q + 2 + 1) = (3q + 3) = 3(q + 1){/tex}, which is clearly divisible by 3.
In this case,{tex} (n + 1){/tex} is divisible by 3.
Hence, one and only one out of {tex}n, (n + 1){/tex} and {tex}(n + 2){/tex} is divisible by 3.