# Closure property

• Nov 23rd 2012, 12:22 PM
loui1410
Closure property
If $a>1$ and $b>1$, how can I show that also $a \cdot b-a-b+2>1$ ?

I saw one solution using partial derivatives from calculus. Is there another way?
• Nov 23rd 2012, 02:11 PM
chiro
Re: Closure property
Hey loui1410.

Just in future, note that the . in your inequality usually refers to the dot product and not multiplication which is usually represented by the appropriate latex symbol [or the appropriate star convention like *].

The calculus way through optimization in my opinion is the best way because it is systematic, general enough, and a well understood technique of finding minimums, maximums (local and global which will always exist but only one of these exists as opposed to potentially many local ones) and because the math has been carefully worked out by very patient and admirable mathematicians.

If you have a situation where you have a general tool-kit that has been peer-reviewed with the highest level of scrutiny in the realm of pure mathematics, then I would advise you to strongly consider those techniques.
• Nov 24th 2012, 03:59 AM
loui1410
Re: Closure property
Thank you, but apparently there is a much simpler solution:
$a*b-a-b+2 = (a-1)(b-1)+1$ and this is clearly greater than 1.
• Nov 24th 2012, 09:07 AM
Plato
Re: Closure property
Quote:

Originally Posted by loui1410
Thank you, but apparently there is a much simpler solution:
$a*b-a-b+2 = (a-1)(b-1)+1$ and this is clearly greater than 1.

Well clearly $(a-1)(b-1)>0$ so $a*b-a-b+2>1$.