Let R be the relation on $ \mathbb{Z} $ given by $ \{ (x,y):x^2+y=2 \} $ Prove that R is function with domain $ \mathbb{Z} $

Question

Let R be the relation on $ \mathbb{Z} $ given by $ \{ (x,y):x^2+y=2 \} $ Prove that R is  function with domain $ \mathbb{Z} $

helder_edwin · Answer

Z is the integers $\mathbb{Z}$, right?

swissgirl · Answer

yaa i just dont the symbol of it in latex

helder_edwin · Answer

just type \mathbb{Z}

helder_edwin · Answer

well u have
$$ \large y=2-x^2 $$
so for a given x you can find (compute) y

helder_edwin · Answer

now suppose $(x,y_1)$ and $(x,y_2)$ are both in R then
$$ \large y_1=2-x^2 $$
and
$$ \large y_2=2-x^2 $$
but this imples $y_1=y_2$
so R is a function

swissgirl · Answer

what abt proving the component that its domain is in Z?

helder_edwin · Answer

$$ \large y=2-x^2 $$
has no restrictions of any kind (division by zero, roots or logarightms) so its domain is $\mathbb{Z}$

swissgirl · Answer

alrighty. Thanks :DDD

helder_edwin · Answer

u r welcome