Question

Need help with a simple proof.

Show that for any pair of real numbers y and z, floor(y) + floor(z) <= floor(y + z)

Answer 1

\(floor(4.8) + floor(5.7) = 4 + 5\), but \(floor(4.8 + 5.7) = floor(10.5) = 10\)

Answer 2

ah, so you just plug in numbers to prove it. Makes sense. Thank you. :D

Answer 3

what @Anickyan showed was just one example of its truth - that does not constitute a proof

Answer 4

Well, not really, but I was just showing an example, from which you should be able to construct proof.

Answer 5

it might help if you expressed y and z as a combination of an integer part and a fractional part

Answer 6

e.g. let:\[y=y_i+y_f\]where \(y_i\) is the integral part of y and \(y_f\) is its fractional part.

Answer 7

The interger part of the floored sum of the numbers might be greater than the sum of the two floored numbers.

Answer 8

Sort of makes sense.

Answer 9

ok, good - let me know if you need any further help on the proof.

Answer 10

ok, I will work on it for a bit and let you know. Thanks for the help.

Answer 11

yw :)

Answer 12

Hmm, I don't really know where to start =\

Answer 13

ok, if we let:\[y=y_i+y_f\]as I described above, then what do you think will be the value of:\[floor(y)=floor(y_i+y_f)=?\]

Answer 14

remember the floor function just throws away the fractional part of a number

Answer 15

they would be the same?

Answer 16

?

Answer 17

I don't understand the question exactly.

Answer 18

do you understand what the floor function does?

Answer 19

Yeah, it rounds down to the lowest whole number

Answer 20

so what is floor(9.9999)?

Answer 21

9

Answer 22

good

Answer 23

so we can also write this as follows...

Answer 24

9.9999 = 9 + 0.9999

therefore:
floor(9.9999) = floor(9 + 0.9999) = 9

i.e. it equals the integral part of 9.9999.

so, using the same technique, what do you think would be the value of:\[floor(y_i+y_f)=?\]

Answer 25

yi

Answer 26

perfect!

Answer 27

so, similarly, what would this equal:\[floor(z)=floor(z_i+z_f)=?\]

Answer 28

zi

Answer 29

great!, so now we know we can write:\[floor(y)+floor(z)=floor(y_i+y_f)+floor(z_i+z_f)=y_i+z_i\]agreed?

Answer 30

Looks good

Answer 31

ok, next we calculate the floor of y+z as follows:\[floor(y+z)=floor(y_i+y_f+z_i+z_f)\]in this expression, we now \(y_i\) and \(z_i\) are whole integers, so they can be pulled out of the expression to give:\[floor(y+z)=floor(y_i+y_f+z_i+z_f)=y_i+z_i+floor(y_f+z_f)\]agreed?

Answer 32

ah yeah, makes sense

Answer 33

good, now concentrate of what is left in this expression, i.e. \(floor(y_f+z_f)\) - can you think of what the minimum value of this expression must be?

Answer 34

not sure, 0 maybe

Answer 35

0 is correct, because the minimum fractional part of any number is zero

Answer 36

similarly, try to think of what the maximum value of \(floor(y_f+z_f)\) could be?

Answer 37

1

Answer 38

perfect!

so now we know that:\[0\le floor(y_f+z_f)\le1\]agreed?

Answer 39

Yep

Answer 40

good, so if we go back up to our earlier steps, you can se that:\[floor(y+z)=floor(y_i+y_f+z_i+z_f)=y_i+z_i+floor(y_f+z_f)\]\[\qquad\ge y_i+z_i\]as we have shown that \(0\le floor(y_f+z_f)\le 1\).

agreed?

Answer 41

Yep, looks good

Answer 42

great! so last step is to go back to our first step where we showed that:\[floor(y)+floor(z)=y_i+z_i\]and plug this into our result to get:\[floor(y+z)\ge floor(y)+floor(z)\]

Answer 43

I hope you understood all the steps.

Answer 44

wow, amazing :O

Answer 45

It makes sense now, thank you!

Answer 46

yw :)

and it only looks amazing now because you may not be familiar with the techniques, but I am sure that you will be up to speed very soon if you keep practicing! :)

Answer 47

I hope so, this is the second proof I've ever tried.

Answer 48

do you have any books or resources you could suggest? :O

Answer 49

the main thing to do with proof is to always first STOP and think about the problem a little - try to think about it in logical terms - it is difficult at first but the more you practice, the better you will get. :)

Answer 50

It's hard because my brain doesn't like to think that way.

Answer 51

never underestimate yourself my friend - you can and you WILL be able to master these soon - just keep at it.

Unfortunately I do not have any books or resources on proofs - I just enjoy maths and have tried to keep up with the subject by teaching and learning at sites like this one. :)

Answer 52

ah cool, well thanks again! :D

Answer 53

ok - and good luck! :)