Question

anybody can help with question number 4 of this assignment : http://imgur.com/qfcdnaf? DISCRETE MATH!!!

Answer 1

you want to do 4a)

Answer 2

For all X for all y there exists z such that x^2 + y^2 -z = 0 .

Answer 3

you can add a colon : for 'such that'

Answer 4

\[(\forall x \in \mathbb{Z^+}) ( \forall y \in \mathbb{Z^+}) (\exists z \in \mathbb{Z^+}:x^2 +y^2-z = 0 )\]

Answer 5

Sorry wasn't on the computer, would you mind explaining your reasoning? Why do you need to explain it is always an element of z^+ ?

Answer 6

well, that is to be explicit what x , y, z are elements of. If you assume the domain is Z+ , then i guess you dont need to use that

Answer 7

maybe @ganeshie8 can look at this . you can also use parentheses like so

Answer 8

\[(\forall x \in \mathbb{Z^+}) ( \forall y \in \mathbb{Z^+}) (\exists z \in \mathbb{Z^+})(x^2 +y^2-z = 0 )\]

Answer 9

if you can assume the universe is Z+ , then you can write it as
\[(\forall x) ( \forall y) (\exists z)\ \ x^2 +y^2-z = 0 \]

Answer 10

Could I technically factor z out? gettubg (∀x∀y∃z)∈Z+?

Answer 11

\[\forall x \ \ \forall y \ \ \exists z \ \ : x^2 +y^2-z = 0 \]

Answer 12

your expression says
"for all x, for all y, there exists a z such (blank) is a member of Z + "
thats not what you want

Answer 13

typo*
your expression says
"for all x, for all y, there exists a z such that (blank) is a member of Z + "

Answer 14

you can combine the for all x for all y i think \[(\forall x,y \in \mathbb{Z^+}) (\exists z \in \mathbb{Z^+})(x^2 +y^2-z = 0 ) \]

Answer 15

i think for reading purposes its better to be explicit . you should state what your domain is explicitly. If it is obvious you are talking about real numbers , then you don't need to state your domain. so for example, you may not need to state your domain here:
\[\forall x \forall y \ \ (x+y = y + x) \]

Answer 16

thats the commutative law for addition