Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a-3b)/b
Now divide by 2, we get
√5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.
Hence, 3 + 2√5 is a irrational number.

(i) Let take that 1/√2 is a rational number.
So we can write this number as
1/√2 = a/b
Here a and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get
1 = (a√2)/b
Now multiply by b
b = a√2
divide by a we get
b/a = √2
Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts.
Hence, 1/√2 is a irrational number


(ii) Let take that 7√5 is a rational number.
So we can write this number as
7√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get
√5 = a/(7b)
Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts.
Hence, 7√5 is a irrational number.


(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
√2 = a/b – 6
√2 = (a-6b)/b
Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number.

But √2 is a irrational number so it contradicts.
Hence, 6 + √2 is a irrational number.

(i) 13/3125
Factorize the denominator we get
3125 =5 × 5 × 5 × 5 × 5 = 5⁵

So denominator is in form of 5^m so it is terminating .

(ii) 17/8
Factorize the denominator we get
8 =2 × 2 × 2 = 2³
So denominator is in form of 2^m so it is terminating .

(iii) 64/455
Factorize the denominator we get
455 =5 × 7 × 13
There are 7 and 13 also in denominator so denominator is not in form of 2^m × 5ⁿ . so it is not terminating.

(iv) 15/1600
Factorize the denominator we get
1600 =2 × 2 × 2 ×2 × 2 × 2 × 5 × 5 = 2⁶ × 5²
so denominator is in form of 2^m × 5ⁿ
Hence it is terminating.

(v) 29/343
Factorize the denominator we get
343 = 7 × 7 × 7 = 7³
There are 7 also in denominator so denominator is not in form of 2^m × 5ⁿ
Hence it is non-terminating.

(vi) 23/(2³ × 5²)
Denominator is in form of 2^m × 5ⁿ
Hence it is terminating.

(vii) 129/(2² × 5⁷ × 7⁵ )
Denominator has 7 in denominator so denominator is not in form of 2^m × 5ⁿ
Hence it is none terminating.

(viii) 6/15
divide nominator and denominator both by 3 we get 2/5
Denominator is in form of 5^m so it is terminating.

(ix) 35/50 divide denominator and nominator both by 5 we get 7/10
Factorize the denominator we get
10=2 × 5
So denominator is in form of 2^m × 5ⁿ so it is terminating.

(x) 77/210
simplify it by dividing nominator and denominator both by 7 we get 11/30
Factorize the denominator we get
30=2 × 3 × 5
Denominator has 3 also in denominator so denominator is not in form of 2^m × 5ⁿ

Hence it is none terminating

(i) 13/3125 = 13/5⁵ = 13×2⁵/5⁵×2⁵ = 416/10⁵ = 0.00416 (ii) 17/8 = 17/2³ = 17×5³/2³×5³ = 17×5³/10³ = 2125/10³ = 2.125

(iv) 15/1600 = 15/2⁴×10² = 15×5⁴/2⁴×5⁴×10² = 9375/10⁶ = 0.009375

(vi) 23/2³5² = 23×5³×2²/2³ 5²×5³×2² = 11500/10⁵ = 0.115

(viii) 6/15 = 2/5 = 2×2/5×2 = 4/10 = 0.4

(ix) 35/50 = 7/10 = 0.7

(i) Since this number has a terminating decimal expansion, it is a rational number of the form p/q, and q is of the form 2^m × 5ⁿ.

(ii) The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.
(iii) Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form p/q, and q is not of the form 2^m × 5ⁿ.