# algebraic expressions

• Sep 3rd 2009, 05:30 PM
pklanni
algebraic expressions
I am a dad helping with daughters homework and have a symbol I have not seen, the symbol is a solid diamond. The problem reads
Quote:

A new operation is defined as a ( solid diamond ) b = a+a-b. Is the set of whole numbers closed under the operation ( solid diamond )? If not, give a set that is closed under this operation.
Can anyone tell me what I am supposed to do with the solid diamond? Thank you in advance.
• Sep 3rd 2009, 05:49 PM
artvandalay11
a (diamond) b=a+a-b

5 (diamond) 2=5+5-2=8

The set of whole numbers is not going to be closed under this new operation (which is simply a combination of familiar operations in addition and subtraction) because the set of whole numbers is not closed under addition and subtraction

Here I am assuming the set of whole numbers is {0,1,2,3,.....}

however the set of integers and real numbers will be closed. The set of whole numbers are not closed because if b is large enough, a+a-b could be negative, and thus the result would not be an element of the whole numbers
• Sep 3rd 2009, 05:52 PM
pickslides
You are given \$\displaystyle a\diamond b = a+a-b \$

So now think, if a = 2 and b = 3 now

\$\displaystyle 2\diamond 3 = 2+2-3 = 1\$

The diamond symbol is not an operation in particular, just defined for this particualr problem.

Even easier

\$\displaystyle a\diamond b = a+a-b = 2a-b\$