Math, asked by worldwaterpark420, 1 month ago

2. Show that any positive odd integer is of the form 6q +1,or 6q+3,6q+5, where q is some integer

Answers

Answered by rhythmarora200
0

Answer:

Step-by-step ex(i) Let x=0.  

47

. Then x=0.474747....

Since two digits are repeating, multiplying both sides by 100, we get

100x=47.474747...=47+0.474747...=47+x

99x=47

x=  

99

47

 

∴0.  

47

=  

99

47

 

(ii) Let x=0.  

001

. Then x=0.001001001...

Since three digits are repeating, multiplying both sides by 1000, we get

1000x=1.001001001...=1+0.001001001...=1+x

1000x−x=1

999x=1

x=  

999

1

 

∴0.  

001

=  

999

1

 

(iii) Let x=0.5  

7

. Then x=0.57777....

Multiplying both sides by 10, we get

10x=5.7777....=5.2+0.57777...=5.2+x

9x=5.2

x=  

9

5.2

 

x=  

90

52

 

∴0.5  

7

=  

90

52

=  

45

26

 

(iv) Let x=0.2  

45

. Then x=0.2454545...

Multiplying both sides by 100, we get

100x=24.545454...=24.3+0.2454545...=24.3+x

99x=24.3

x=  

99

24.3

 

0.2  

45

=  

990

243

=  

110

27

 

(v) Let x=0.  

6

. Then x=0.66666...

Multiplying both sides by 10, we get

10x=6.66666....=6+0.6666...=6+x

9x=6

x=  

9

6

=  

3

2

 

∴0.  

6

=  

3

2

 

(vi) Let x=1.  

5

. Then x=1.55555...

Multiplying both sides by 10, we get

10x=15.5555...=14+1.5555...=114+x

9x=14

x=  

9

14

 

∴1.  

5

=1  

9

5

 

So, every number with a non-terminating and recurring decimal expansion can be expressed in the form  

q

p

, where p and q are integers and q not equal to zero.planation:

Answered by mehulkumarvvrs
0

Using Euclid division algorithm, we know that a = bq + r, 0 ≤ r ≤ b ----(1)

Let, a be any positive integer and b = 6.

Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5  

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

6q + 0 : 6 is divisible by 2, so it is an even number.

6q + 1 : 6 is divisible by 2, but 1 is not divisible by 2 so it is an odd number.

6q + 2 : 6 is divisible by 2, and 2 is divisible by 2 so it is an even number.

6q + 3 : 6 is divisible by 2, but 3 is not divisible by 2 so it is an odd number.

6q + 4 : 6 is divisible by 2, and 4 is divisible by 2 so it is an even number.

6q + 5 : 6 is divisible by 2, but 5 is not divisible by 2 so it is an odd number.

And therefore, any odd integer can be expressed in the form 6q + 1 or 6q + 3 or 6q+5.

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