Bergman Rolf

answers: 1

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Answer:

Answer:

Theorem: Let x=

q

p

be a rational number, such that the prime factorisation of q is of the form 2

n

5

m

, where n, m are non-negative integers. Then, x has a decimal expansion which terminates.

(i)

3125

13

Factorise the denominator, we get

3125=5×5×5×5×5=5

5

So, denominator is in form of 5

m

so,

3125

13

is terminating.

(ii)

8

17

Factorise the denominator, we get

8=2×2×2=2

3

So, denominator is in form of 2

n

so,

8

17

is terminating.

(iii)

455

64

Factorise the denominator, we get

455=5×7×13

So, denominator is not in form of 2

n

5

m

so,

455

64

is not terminating.

(iv)

1600

15

Factorise the denominator, we get

1600=2×2×2×2×2×2×5×5=2

6

5

2

So, denominator is in form of 2

n

5

m

so,

1600

15

is terminating.

(v)

343

29

Factorise the denominator, we get

343=7×7×7=7

3

So, denominator is not in form of 2

n

5

m

so,

343

29

is not terminating.

(vi)

2

3

5

2

23

Here, the denominator is in form of 2

n

5

m

so,

2

3

5

2

23

is terminating.

(vii)

2

2

5

7

7

5

129

Here, the denominator is not in form of 2

n

5

m

so,

2

2

5

7

7

5

129

is not terminating.

(viii)

15

6

Divide nominator and denominator both by 3 we get

15

3

So, denominator is in form of 5

m

so,

15

6

is terminating.

476

Narn

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