Question

1)Prove the statement is true
2)Is the converse true? if so, prove it
3)Is the contrapositive true? if so, prove it
4)Is the negation true? if so, prove it

Answer 1

for all real numbers x and y, if \[x -\lfloor x \rfloor< \frac{ 1 }{ 2} \]
and \[y -\lfloor y \rfloor< \frac{ 1 }{ 2} \]
then \[\lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor\]

Answer 2

[...] define the greatest integer right ?

Answer 3

umm i dont know

Answer 4

the statement is above, the question doesnt say find the greatest integer, it asks to prove that the above is true if you assume x and y are real numbers

Answer 5

we have an If P then Q statement... and that's all I could give :(

Answer 6

thanks...all i have is suppose x,y are reals, suppose P is true

Answer 7

dont know where to start

Answer 8

do you know the definitions and axioms?

Answer 9

ok its like this
if u have y=3.18..
then [y]=4 ok ?

Answer 10

these proof crud things are like well you need to apply the right definition and then think outside the dam n box

Answer 11

I'm studying it now, but my class is doing induction

Answer 12

just finished the Tower of Hanoi

Answer 13

yes i have definitions for floor and ceilings, but dunno how i would use them, it would introduce more variables and etc

Answer 14

x-[x]<0.5 is true
y-[y]<0.5 is true
proof
[x+y]=[x]+[y]

Answer 15

brb ok ?

Answer 16

k

Answer 17

the negation would be everything negative on this statement

Answer 18

i just know my instructor said something about adding the left side of the inequalities together

Answer 19

the negation is those two equations you got are greater than 1/2

Answer 20

and the end result is that they shouldn't equal each other

Answer 21

x−⌊x⌋ + y−⌊y⌋ < 1...i think

Answer 22

O_O

Answer 23

x + y < 1 + ⌊x⌋ + ⌊y⌋

Answer 24

i dunno, lol

Answer 25

:O no :_:

Answer 26

add first two given conditions

Answer 27

1 + ⌊x⌋ + ⌊y⌋ is the ceiling of x+y?

Answer 28

\(x -\lfloor x \rfloor< \frac{ 1 }{ 2} \)

\(y -\lfloor y \rfloor< \frac{ 1 }{ 2} \)

Answer 29

and thus it's floor is ⌊x⌋ + ⌊y⌋ ..

Answer 30

adding,

\(\large x + y - \lfloor x \rfloor - \lfloor y \rfloor < 1 \)

Answer 31

\(\large x + y < 1 + \lfloor x \rfloor + \lfloor y \rfloor \)

Answer 32

now take the floor both sides

Answer 33

\(\large \lfloor x + y \rfloor < \lfloor 1 + \lfloor x \rfloor + \lfloor y \rfloor \rfloor \)

Answer 34

\[\lfloor x \rfloor + \lfloor y \rfloor \le x + y < \lfloor x \rfloor + \lfloor y \rfloor +1\]

Answer 35

floor of x and floor of y gives u the sum of integer values,
and floor of a number less than 1 is simply 0

Answer 36

so,

\(\large \lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor \)

Answer 37

so the given statement is true

Answer 38

can you explain the right side again please?

Answer 39

\(\large \lfloor x + y \rfloor < \lfloor 1 + \lfloor x \rfloor + \lfloor y \rfloor \rfloor \)

take for example : x = 3.4, y = 7.2
[x] = 3
[y] = 7
[<1] = 0

Answer 40

right hand evaluates to 0+3+7 = 10,
which is same as left hand side : [3.4 + 7.2] = [10.6] = 10

Answer 41

that example is oly for you to *see*
thats not required for the main proof..

Answer 42

yeah, im trying to show this without subbing in numbers

Answer 43

'was already shown before the example.. go thru again :P

Answer 44

⌊1+⌊x⌋+⌊y⌋⌋ to ⌊x⌋+⌊y⌋ im still confused about this

Answer 45

if u digest below, it wud make sense :-

if 0 < a < 1, then
[a] = 0

Answer 46

[0.99] = [0.1] = [0.4] = 0

Answer 47

see if below argument makes sense :-

say x-[x] = a
and y-[y] = b

then from the hypothesis, a+b < 1

adding the given statements :
x+y = a+b + [x] + [y]

taking floor both sides :

[x+y] = [ a+b + [x] + [y]]
= [(a+b) + integer]
= [a+b] + integer
= 0 + integer
= [x] + [y]