anonymous
  • anonymous
Need help with a simple proof. Show that for any pair of real numbers y and z, floor(y) + floor(z) <= floor(y + z)
Discrete Math
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\(floor(4.8) + floor(5.7) = 4 + 5\), but \(floor(4.8 + 5.7) = floor(10.5) = 10\)
anonymous
  • anonymous
ah, so you just plug in numbers to prove it. Makes sense. Thank you. :D
asnaseer
  • asnaseer
what @Anickyan showed was just one example of its truth - that does not constitute a proof

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anonymous
  • anonymous
Well, not really, but I was just showing an example, from which you should be able to construct proof.
asnaseer
  • asnaseer
it might help if you expressed y and z as a combination of an integer part and a fractional part
asnaseer
  • 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
  • anonymous
The interger part of the floored sum of the numbers might be greater than the sum of the two floored numbers.
anonymous
  • anonymous
Sort of makes sense.
asnaseer
  • asnaseer
ok, good - let me know if you need any further help on the proof.
anonymous
  • anonymous
ok, I will work on it for a bit and let you know. Thanks for the help.
asnaseer
  • asnaseer
yw :)
anonymous
  • anonymous
Hmm, I don't really know where to start =\
asnaseer
  • 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
  • asnaseer
remember the floor function just throws away the fractional part of a number
anonymous
  • anonymous
they would be the same?
asnaseer
  • asnaseer
?
anonymous
  • anonymous
I don't understand the question exactly.
asnaseer
  • asnaseer
do you understand what the floor function does?
anonymous
  • anonymous
Yeah, it rounds down to the lowest whole number
asnaseer
  • asnaseer
so what is floor(9.9999)?
anonymous
  • anonymous
9
asnaseer
  • asnaseer
good
asnaseer
  • asnaseer
so we can also write this as follows...
asnaseer
  • 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
  • anonymous
yi
asnaseer
  • asnaseer
perfect!
asnaseer
  • asnaseer
so, similarly, what would this equal:\[floor(z)=floor(z_i+z_f)=?\]
anonymous
  • anonymous
zi
asnaseer
  • 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
  • anonymous
Looks good
asnaseer
  • 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
  • anonymous
ah yeah, makes sense
asnaseer
  • 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
  • anonymous
not sure, 0 maybe
asnaseer
  • asnaseer
0 is correct, because the minimum fractional part of any number is zero
asnaseer
  • asnaseer
similarly, try to think of what the maximum value of \(floor(y_f+z_f)\) could be?
anonymous
  • anonymous
1
asnaseer
  • asnaseer
perfect! so now we know that:\[0\le floor(y_f+z_f)\le1\]agreed?
anonymous
  • anonymous
Yep
asnaseer
  • 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
  • anonymous
Yep, looks good
asnaseer
  • 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
  • asnaseer
I hope you understood all the steps.
anonymous
  • anonymous
wow, amazing :O
anonymous
  • anonymous
It makes sense now, thank you!
asnaseer
  • 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
  • anonymous
I hope so, this is the second proof I've ever tried.
anonymous
  • anonymous
do you have any books or resources you could suggest? :O
asnaseer
  • 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
  • anonymous
It's hard because my brain doesn't like to think that way.
asnaseer
  • 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
  • anonymous
ah cool, well thanks again! :D
asnaseer
  • asnaseer
ok - and good luck! :)

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