Check if ive done right please binary operations

**adam_leeds** · January 7th 2010, 11:40 AM

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

**Drexel28** · January 7th 2010, 01:08 PM

Originally Posted by adam_leeds

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 $*:S^2\mapsto S$ so when $S=\mathbb{N}$ we ave that $*(3,4)\notin\mathbb{N}$ . But, why doesn't that exact same example work when $S=\mathbb{Q}^+$ ?

**hatsoff** · January 7th 2010, 01:15 PM

Originally Posted by adam_leeds

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

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

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

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

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

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

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

**Drexel28** · January 7th 2010, 01:25 PM

Originally Posted by hatsoff

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

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

I'm sorry. Did I say something incorrect?

**hatsoff** · January 7th 2010, 01:52 PM

Originally Posted by Drexel28

I'm sorry. Did I say something incorrect?

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.

**adam_leeds** · January 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?

**Swlabr** · January 8th 2010, 06:58 AM

Originally Posted by adam_leeds

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:

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

$4 = \frac{4}{1}$ .

**adam_leeds** · January 8th 2010, 07:03 AM

Originally Posted by Swlabr

3 and 4 are fractions:

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

$4 = \frac{4}{1}$ .

woops

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