Question

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

Answer 1

Z is the integers \(\mathbb{Z}\), right?

Answer 2

yaa i just dont the symbol of it in latex

Answer 3

just type \mathbb{Z}

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

alrighty. Thanks :DDD

Answer 9

u r welcome