sir pls answer
Given A = 0.a1a2a3a1a2a3.......
and B = 0.b1b2b1b2.......
To Prove: 10989 × (a + b) is an integer.
Proof: First represent A and B in the form of rational number.
We have, A = 0.a1a2a3 a1a2a3.......(1)
Here, we have three repeating digits after the decimal point. So, we multiply both sides of (1) by 1000 to get
1000 A = a1a2a3·a1a2a3 a1a2a3....... (2)
Now, (2)–(1), will give
999A = a1a2a3
Again, we have, B =0.b1b2b1b2........ (3)
Here, we have two repeating digits after the decimal point. So, we multiply both sides of (3) by 100 to get
100 B = b1b2·b1b2 b1b2 ........ (4)
Now, (4)–(3), will give,
99B = b1b2
It is given that a1,a2,a3,b1 and b2 are integers between 1 to 9.
∴ From (5), it is clear that, 10989 (A + B) is an integer.