# Check if ive done right please binary operations

• Jan 7th 2010, 11:40 AM
Check if ive done right please binary operations
An operation * is defined on a set of numbers S by x * y = x + y - 2(x^2)(y^2)

in z+, q and q+

for z + its not a binary operation because you can get a negative answer say x = 3 and y = 4

for q it is a binary operation

for q+ it is a binary operation
• Jan 7th 2010, 01:08 PM
Drexel28
Quote:

An operation * is defined on a set of numbers S by x * y = x + y - 2(x^2)(y^2)

in z+, q and q+

for z + its not a binary operation because you can get a negative answer say x = 3 and y = 4

for q it is a binary operation

for q+ it is a binary operation

You are correct to note that $\displaystyle *:S^2\mapsto S$ so when $\displaystyle S=\mathbb{N}$ we ave that $\displaystyle *(3,4)\notin\mathbb{N}$. But, why doesn't that exact same example work when $\displaystyle S=\mathbb{Q}^+$?
• Jan 7th 2010, 01:15 PM
hatsoff
Quote:

An operation $\displaystyle *$ is defined on a set of numbers $\displaystyle S$ by $\displaystyle x * y = x + y - 2x^2y^2$

in $\displaystyle \mathbb{Z}^+$, $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{Q}^+$

for $\displaystyle \mathbb{Z}^+$ its not a binary operation because you can get a negative answer say $\displaystyle x = 3$ and $\displaystyle y = 4$

for $\displaystyle \mathbb{Q}$ it is a binary operation

for $\displaystyle \mathbb{Q}^+$ it is a binary operation

Since $\displaystyle (3,4)\in\mathbb{Q}^+\times\mathbb{Q}^+$, then clearly $\displaystyle \mathbb{Q}^+\times\mathbb{Q}^+$ is not a valid domain for the above function into $\displaystyle \mathbb{Q}^+$.

However, $\displaystyle \mathbb{Q}\times\mathbb{Q}$ is a valid domain into $\displaystyle \mathbb{Q}$.
• Jan 7th 2010, 01:25 PM
Drexel28
Quote:

Originally Posted by hatsoff
Since $\displaystyle (3,4)\in\mathbb{Q}^+\times\mathbb{Q}^+$, then clearly $\displaystyle \mathbb{Q}^+\times\mathbb{Q}^+$ is not a valid domain for the above function into $\displaystyle \mathbb{Q}^+$.

However, $\displaystyle \mathbb{Q}\times\mathbb{Q}$ is a valid domain into $\displaystyle \mathbb{Q}$.

I'm sorry. Did I say something incorrect? (Worried)
• Jan 7th 2010, 01:52 PM
hatsoff
Quote:

Originally Posted by Drexel28
I'm sorry. Did I say something incorrect? (Worried)

Oh no. It just takes me longer to type than you, apparently, so that I logged my response after yours even though we set out at approximately the same time.
• Jan 8th 2010, 06:11 AM
Sorry i dont understand why q+ is not a binary operation. I thought q meant fractions. Why are you using 3 and 4?
• Jan 8th 2010, 06:58 AM
Swlabr
Quote:

Sorry i dont understand why q+ is not a binary operation. I thought q meant fractions. Why are you using 3 and 4?

3 and 4 are fractions:

$\displaystyle 3 = \frac{3}{1}$,

$\displaystyle 4 = \frac{4}{1}$.
• Jan 8th 2010, 07:03 AM
$\displaystyle 3 = \frac{3}{1}$,
$\displaystyle 4 = \frac{4}{1}$.