Need help with a simple proof.
Show that for any pair of real numbers y and z, floor(y) + floor(z) <= floor(y + z)

- anonymous

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- anonymous

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

- anonymous

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

- asnaseer

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

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- anonymous

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

- asnaseer

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

- asnaseer

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

- anonymous

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

- anonymous

Sort of makes sense.

- asnaseer

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

- anonymous

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

- asnaseer

yw :)

- anonymous

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

- asnaseer

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

- asnaseer

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

- anonymous

they would be the same?

- asnaseer

?

- anonymous

I don't understand the question exactly.

- asnaseer

do you understand what the floor function does?

- anonymous

Yeah, it rounds down to the lowest whole number

- asnaseer

so what is floor(9.9999)?

- anonymous

9

- asnaseer

good

- asnaseer

so we can also write this as follows...

- asnaseer

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:\[floor(y_i+y_f)=?\]

- anonymous

yi

- asnaseer

perfect!

- asnaseer

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

- anonymous

zi

- asnaseer

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?

- anonymous

Looks good

- asnaseer

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?

- anonymous

ah yeah, makes sense

- asnaseer

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?

- anonymous

not sure, 0 maybe

- asnaseer

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

- asnaseer

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

- anonymous

1

- asnaseer

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

- anonymous

Yep

- asnaseer

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?

- anonymous

Yep, looks good

- asnaseer

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:\[floor(y+z)\ge floor(y)+floor(z)\]

- asnaseer

I hope you understood all the steps.

- anonymous

wow, amazing :O

- anonymous

It makes sense now, thank you!

- asnaseer

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! :)

- anonymous

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

- anonymous

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

- asnaseer

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. :)

- anonymous

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

- asnaseer

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. :)

- anonymous

ah cool, well thanks again! :D

- asnaseer

ok - and good luck! :)

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