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\(floor(4.8) + floor(5.7) = 4 + 5\), but \(floor(4.8 + 5.7) = floor(10.5) = 10\)
ah, so you just plug in numbers to prove it. Makes sense. Thank you. :D
what @Anickyan showed was just one example of its truth - that does not constitute a proof
Well, not really, but I was just showing an example, from which you should be able to construct proof.
it might help if you expressed y and z as a combination of an integer part and a fractional part
e.g. let:\[y=y_i+y_f\]where \(y_i\) is the integral part of y and \(y_f\) is its fractional part.
The interger part of the floored sum of the numbers might be greater than the sum of the two floored numbers.
Sort of makes sense.
ok, good - let me know if you need any further help on the proof.
ok, I will work on it for a bit and let you know. Thanks for the help.
Hmm, I don't really know where to start =\
ok, if we let:\[y=y_i+y_f\]as I described above, then what do you think will be the value of:\[floor(y)=floor(y_i+y_f)=?\]
remember the floor function just throws away the fractional part of a number
they would be the same?
I don't understand the question exactly.
do you understand what the floor function does?
Yeah, it rounds down to the lowest whole number
so what is floor(9.9999)?
so we can also write this as follows...
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:\[floor(y_i+y_f)=?\]
so, similarly, what would this equal:\[floor(z)=floor(z_i+z_f)=?\]
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?
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?
ah yeah, makes sense
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?
not sure, 0 maybe
0 is correct, because the minimum fractional part of any number is zero
similarly, try to think of what the maximum value of \(floor(y_f+z_f)\) could be?
perfect! so now we know that:\[0\le floor(y_f+z_f)\le1\]agreed?
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?
Yep, looks good
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:\[floor(y+z)\ge floor(y)+floor(z)\]
I hope you understood all the steps.
wow, amazing :O
It makes sense now, thank you!
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! :)
I hope so, this is the second proof I've ever tried.
do you have any books or resources you could suggest? :O
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. :)
It's hard because my brain doesn't like to think that way.
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. :)
ah cool, well thanks again! :D
ok - and good luck! :)