11thanks Sir [1]
1)Modulus Function
if a, b are positive real numbers,then
x2 ≥ a22 <=> lxl ≥ a <=> x ≤ (-a) or x ≥ a.
HOW x ≤ (-a) ?
2) Greatest Integer Function
[x] > k => x ≥ k + 1, where k is an integer. Please prove this with some example, am not able to understand this property.
3)Smallest integer function.
(-x) = -(x) + 1, where x belongs to R-Z, prove this with some example.
I am not able to make the bracket which are used to represent smallest integer functions, so I made usage of open bracket.
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5 Answers
62x2>a2
(x-a)(x+a)>0
so x-a > 0 and x+a>0 or both are greater than 0
so x>a and x>-a or otherwise x< a and x<-a
if a is positive then
x>a and x>-a can be summarized as x>a>-a
while
x< a and x<-a can be summarized as x< a<-a
I hope this makes the first part clear :)
622) Greatest Integer Function
[x] > k => x ≥ k + 1, where k is an integer. Please prove this with some example, am not able to understand this property.
all that this says is that if the greatest integer is greater than another integer then the number is greater than equal to k+1
a simple proof will be that x-1<[x]<=x
we also know that [x]>k>x-1
so k+1>x
1[ x ] is an integer and k is also integer ... and [ x ] > k ....so minimum difference between them can be =1 or > 1 ....
hence [ x ] >= k + 1
13) ( - x) + (x) = 1
1)when x belongs to (0,1)
0 + 1 =1 (proved)
2) when x belongs to (-1 ,0)
1 + 0 =1 (proved)
3)when x does not belong to (0,1) but is +
let it be m
then (-m) + ( m) =1
(- m) will be (-) integer and (m) will be a (+) integer
(-m) goes to - integer< m
(m) goes to + integer > m
so (m) >(-m) ( proved)
(m ) + ( -m ) > 0
as both are intezers
so (m) + (-m) >= 1
now u can prove by taking examples that >1 case does not arise
similarly another case can be proved when x does not belongs to (-1,0) and is - ..............................................................................................