For the 2nd one, think of it like this:
where k is an integer, and p is a non-integer between 0 and 1.
In mathematical terms:
Hence:
On the RHS, we know that k is the integer part of the number so:
In other words, the NON-integer part of the LHS must be
So now you need to come up with values of k and p such that:
For example, p = 0.1, k = 1 is a solution. (x = 1.1, in other words).
i'm not sure about 1) but i guess there is no solution. my solutions to 2) and 3):
clearly any integer is a solution. so we'll assume that let then and which is equivalent to
case 1. : in this case, since we'll have: but from (1) we have this contradiction means there's no solution in this case.
case 2. : then the conditions and (1) are equivalent to conversely if for some positive integer then
and thus and
so the complete set of solutions of the equation is:
the equation is equivalent to these two conditions together: and
3)
let where solving the inequalities in (2) for gives you: which gives us thus
There is no solution
Using the following two properties:
It can be deduced that:
Hence:
And in general
Continuing like this, the equation can be reduced to:
It is non-integer, and so there are no solutions.
look this
When he plugs , the result is .We're going to need to take off three. Now imagine we decrease that slightly. Every term on the lefthand side will decrease by AT LEAST one (for example, ), therefore the sum will decrease by AT LEAST six, and that's too much.
some one can more explain this way