I could be wrong but I would say b)1. This is because here three cases can be considered 1. Both x and y are integers 2. Both x and y have o.5 as the decimal part. 3. One of these is an integer and the other has a decimal part of 0.5 In each of these cases the difference between 2x and 2y will have to be covered by the difference between [x] and [y], which is not possible even by taking minimum difference between x any y in each of the cases mentioned above.
I have not given the correct reasoning in my first response to this problem.This is what I mean. Firstly, we assign values to x any y to get a values of a and b.Now we have to find a pair of x and y for which the values of a and b are the same.for this four cases are possible 1. If x,y are integers and the next values taken are integers with a min. difference of 1. In this case. [x] + 2y =a If both x and y are being integers are increased or decreased by 1, a cannot be obtained again. 2.If x,y are both have 0.5 as the decimal part. Two cases are possible here a)If x and y are increased here lets say by 0.5 Then [x] and [y] will increase to the next integer and so will the value of 2x and 2y, so again obtaining a and b from the equations will not be possible. b) If decreased by 0.5 the [x],[y] remain same but again 2x and 2y change. 3.if one of x and y has o.5 as the decimal part and the other is an integer, in this case as well the integer part will increase or decrease by 1 and will be doubled on multiplying by2.and the [x],[y]will be increased to the next integer or remain same depending on the next value chosen is greater or less. Hence there b)1
I would proceed like this.. say [x] = greatest integer <=x and {x} = fractional part of x.
For positive integer So x = [x] + {x}.
Please see its not very difficult to observe that for negative x. x = [x] + 1 + {x}.
Consider the case where both x and y are positive. then 1st equation can be written as:- [x] + 2[y] + 2{y} = a ........ (Equ 1') and second equation can be rewritten as :- [y] + 2[x] + 2{x} = b ............(Equ 2')
Now 1' + 2' gives:- 3[x] + 3[y] + 2({x} + {y}) = a + b ....... Equ 3'
and 1' - 2' gives [y] - [x] + 2({y} - {x}) = a - b ..........Equ 4'
since {x} + {y} can take values 0, 0.5, 1.5 in order to satisfy 3' and {x} - {y} can take values 0, 0.5 in order to satisy 4'
I could be wrong but I would say b)1.
ReplyDeleteThis is because here three cases can be considered
1. Both x and y are integers
2. Both x and y have o.5 as the decimal part.
3. One of these is an integer and the other has a decimal part of 0.5
In each of these cases the difference between 2x and 2y will have to be covered by the difference between [x] and [y], which is not possible even by taking minimum difference between x any y in each of the cases mentioned above.
I have not given the correct reasoning in my first response to this problem.This is what I mean.
ReplyDeleteFirstly, we assign values to x any y to get a values of a and b.Now we have to find a pair of x and y for which the values of a and b are the same.for this four cases are possible
1. If x,y are integers and the next values taken are integers with a min. difference of 1.
In this case.
[x] + 2y =a
If both x and y are being integers are increased or decreased by 1, a cannot be obtained again.
2.If x,y are both have 0.5 as the decimal part.
Two cases are possible here
a)If x and y are increased here lets say by 0.5 Then [x] and [y] will increase to the next integer and so will the value of 2x and 2y, so again obtaining a and b from the equations will not be possible.
b) If decreased by 0.5 the [x],[y] remain same but again 2x and 2y change.
3.if one of x and y has o.5 as the decimal part and the other is an integer, in this case as well the integer part will increase or decrease by 1 and will be doubled on multiplying by2.and the [x],[y]will be increased to the next integer or remain same depending on the next value chosen is greater or less.
Hence there b)1
Pardon me for my mistakes. I am fallible!!
Nice attempt, but you have misunderstood the problem !
ReplyDeleteI would proceed like this..
ReplyDeletesay [x] = greatest integer <=x
and {x} = fractional part of x.
For positive integer
So x = [x] + {x}.
Please see its not very difficult to observe that for negative x.
x = [x] + 1 + {x}.
Consider the case where both x and y are positive.
then 1st equation can be written as:-
[x] + 2[y] + 2{y} = a ........ (Equ 1')
and second equation can be rewritten as :-
[y] + 2[x] + 2{x} = b ............(Equ 2')
Now 1' + 2' gives:-
3[x] + 3[y] + 2({x} + {y}) = a + b ....... Equ 3'
and 1' - 2' gives
[y] - [x] + 2({y} - {x}) = a - b ..........Equ 4'
since {x} + {y} can take values 0, 0.5, 1.5 in order to satisfy 3'
and {x} - {y} can take values 0, 0.5 in order to satisy 4'
Now there are fe