the-range-of-y-x-is-0-if-x-0-or-just-0-

Question Number 90457 by M±th+et£s last updated on 23/Apr/20

t h e r a n g e o f y = \sqrt{x} i s [0, + \infty) i f x ⩾ 0

o r j u s t [0, + \infty) ?

Commented by MJS last updated on 24/Apr/20

talking of “ {definition}^{″} and “ {range}^{″} makes

no sense for x \notin R usually .

y = \sqrt{x} is defined for x ⩾ 0 and its range is [0; + \infty)

for x \in C we get different kinds of problems

[try to solve - 1 = \sqrt{x}]

Commented by M±th+et£s last updated on 24/Apr/20

t h a t m e a n s t h e r a n g e i s [0, + \infty) b e c a u s e

t h e r a n g e i s y v a l u e i n R o r i t s i n R a n d C ?

a n d i w a s a s k i n g i f i t s w r o n g i f w e {d i d n}^{'} t

w r o t e x ⩾ 0

Commented by MJS last updated on 24/Apr/20

first you must specify the underlying set

for x and for y

i . e . f = {(x, y) ∣ x \in N \land y \in N \land y = \sqrt{x}}

or f = {(x, y) ∣ x \in R \land y \in Z \land y = \sqrt{x}}

or whatever is required

usually, after having learned the rules for

N, Z, Q and R you learn about functions

with R \to R or (x, y) \in R \times R (= R^{2})

(x, y) \in C^{2} is the last step

so I guess {we}^{'} re in R^{2}

in this case y = \sqrt{x} is not defined for x < 0 and

the range is [0; + \infty) or simply y ⩾ 0 . if only

the range is asked for, y ⩾ 0 is enough

Commented by MJS last updated on 24/Apr/20

if {we}^{'} re in C^{2} {it}^{'} s getting more complicated

x \in C \Leftrightarrow x = e^{i θ} r with r \in R^{+} and - π ⩽ θ < π (?)

here the complications start . some use

0 ⩽ θ < 2 π . some say \sqrt{e^{i θ} r} is not unique because

y^{2} = x has 2 solutions \dots but \sqrt{4} \neq - 2 even though

{(- 2)}^{2} = 4 . which definitions to use depends

on which kind of solution you need .

y = \sqrt{x} = \sqrt{e^{i θ} r} = e^{i \frac{θ}{2}} \sqrt{r} is unique if we define

r ⩾ 0 \land - π ⩽ θ < π

then y = \sqrt{x} is defined for x \in C

but you cannot get y \in R^{-} because

- π ⩽ θ < π \Leftrightarrow - \frac{π}{2} ⩽ \frac{θ}{2} < \frac{π}{2}

\Rightarrow the range is y = e^{i ϕ} s; s \in R^{+} \land - \frac{π}{2} ⩽ ϕ < \frac{π}{2}

or y = a + b i; a \in R^{+} \land b \in R

change the definitions above and you get

different ranges

Commented by M±th+et£s last updated on 24/Apr/20

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