Ceil function proof

Question

I'm sorry if this is a repeated question but I can't find it anywhere. Any help is highly appreciated How to prove that for any +ve real number

ceil (a*b) <= ceil(a)*ceil(b)

Thankyou very much

Answer 1

So first of all from the definition like ceil is the lower bound of set of integers such that ceil(x) is greater than or equal to x

so its valid for all real numbers x

so from the definition ceil(a) ≥ a this also implies ceil(b) ≥ b

so here you said that it is a,b are positive integers so

obviously a*b is an integer this mean ceil(ab)=ab=ceil(a)*ceil(b)

so now lets suppose a,b are two positive reals then let a=ceil(a)-e1 and b=ceil(b)-e2 where e1,e2 are non negative reals less than 1

a*b is obviously less than or equal to ceil(a)*ceil(b) because a <= ceil(a) and b<=ceil(b)

a*b<=ceil(a) * ceil(b)
applying ceil on both sides we get

ceil(a*b) <= ceil( ceil(a)*ceil(b) )
since ceil(a)*ceil(b) is integer ceil( ceil(a)*ceil(b) ) = ceil(a)*ceil(b)
this implies ceil(a*b) <= ceil(a) * ceil(b)

Answer 2

By definition of ceil and since $a,b$ are positive, $ab \le \lceil a \rceil \lceil b \rceil$.

Now $\lceil a \rceil \lceil b \rceil$ is an integer greater than or equal to $ab$, while $\lceil ab \rceil$ is the smallest such integer.

Hence $\lceil ab \rceil \le \lceil a \rceil \lceil b \rceil$.

Answer 3

If $a> 0; b> 0$ then let $M = \{n\in \mathbb N|n \ge ab\}$.

Note $0< a \le \lceil a\rceil$ and $0 < b \le \lceil b\rceil$ so $ab \le \lceil a\rceil\lceil b\rceil$ so $\lceil a\rceil\lceil b\rceil \in M$.

And $ab \le \lceil ab\rceil$ so $\lceil ab\rceil\in M$, but not only is $\lceil ab\rceil\in M$, but $\lceil ab\rceil-1 < ab < \lceil ab\rceil$ so $\lceil ab\rceil = \min M$.

And as $\min M \le k$ for all $k \in M$ we must have $\lceil ab\rceil\le \lceil a\rceil\lceil b\rceil$.

The End.

.... to show that $\lceil a\rceil\lceil b\rceil$ need not be equal to $\lceil ab\rceil$

we can not that $(\lceil a\rceil-1) (\lceil b\rceil-1)=\lceil a\rceil\lceil b\rceil- (\lceil a\rceil+\lceil b\rceil-1) < ab \le \lceil a\rceil\lceil b\rceil$ and that is a range that could (doesn't have to be but could) be a range of many integers.

A good exmample of this would be $a=5.1$ and $b=7.1$ so $ab = 36.21$

We have $5 < a < 6$ and $7< b < 8$ so $35 < ab = 36.21 < 48=\lceil 5.1\rceil \lceil 7.1\rceil$. That is a loose range of $13$ possible integers.

But $36 < ab=36.21 < 37=\lceil 5.1\cdot 7.1\rceil$. That is a tight range of only $1$ possible integer.