Question

g

Answer 1

hii..welcome to openstudy !

Answer 2

Hi :)

Answer 3

ok so do you know [x] when x is an integer ?

Answer 4

yes, it's x itself

Answer 5

so for that cases we get [2x] - 2[x] = 0

agree?

Answer 6

agreed

Answer 7

ok now lets look at other cases

if x is not an integer we get the the integer less than x
ie x-1 for [x]
so 2 times will be 2(x-1) =2x-2

now for [2x] when x is not an integer we will get 2x-1

subtracting

2x-1-(2x-2)
=2x-1-2x+2
=1

:)

Answer 8

does that cover all of R?

Answer 9

of course we considered integers and non integers ,their union is R

Answer 10

yup :D

Answer 11

ok but i think we should do 2nd part in another way

Answer 12

[x] cannot be taken as -1 for non integers x

[2.5]=2 and not 1

Answer 13

*x-1

Answer 14

yup you're right

Answer 15

let me think of an alternate solution \\oh you posted another qn in between !

Answer 16

[3.14] = 3

Answer 17

isnt it better if u had posted one after we finished this ?

Answer 18

Is it not possible to open this again?

Answer 19

nope

Answer 20

oh snap

Answer 21

while i think of an alternate solution check this out :

http://prezi.com/fs3hqdpcopic/an-unofficial-guide-to-openstudy/

Answer 22

ok got it :)

Answer 23

there?

Answer 24

yup

Answer 25

This is what i get now:

[x]=x+1-k if x is not an integer where k is the fractional part
2[x]=2x+2-2k

[2x]=2x+1-2k

therefore
[2x]-2[x]=-1 when x is not an integer


can you check the question again to see if its -1 or +1 ?

Answer 26

+1

Answer 27

then i suppose i am making a mistake somewhere ..let me recheck

Answer 28

did you check for negative numbers

Answer 29

hmm...not yet but our first step is true ;)

Answer 30

can i know where did you get this question from? is it from your textbook?

Answer 31

This question was in one of our assignments.

Answer 32

ok

Answer 33

finally got it :)

Answer 34

what you said was right i didnt consider negative numbers

Answer 35

let me start afresh ....

for x=integer
____________

[2x]-2[x]=0

when x is not an integer we take 2 cases

for x>0
_______

[x]=x-k where k is fractional part

2[x]=2x-2k

now

[2x]=2x-2k

so [2x]-2[x]=0

for x<0
________

[x]=x-1+k where k is fractional part
2[x]=2x-2+2k

[2x]=2x-1+2k

therefore
[2x]-2[x]=2x-1+2k-2x+2-2k
=1

Hence proved

Answer 36

Is that clear ?

Answer 37

yup :)

Answer 38

it would be a good idea to click best response if a user spents his/her time to help you

Answer 39

its the way of thanking in openstudy ! ^_^