Question

Are rational expressions equivalent to their simplified forms?

Answer 1

I am inclined to say yes but I'm not sure as they sometimes have different domains

Answer 2

due to the same reason, I'd say no :)

Answer 3

due to the constraints in the domain

Answer 4

so why does the domain change once you simplify the expression?

Answer 5

well, the domain doesn't really change, for the original expression

Answer 6

sorry, kind of confused. can you explain a little more because I don't want to seem like I don't know what I'm talking about when I answer this LOL

Answer 7

well, the obvious cases are rationals
$$
\cfrac{x-3}{x^2-9} \implies \cfrac{x-3}{(x-3)(x+3)} \implies \cfrac{1}{x+3}
$$
the original constraints, still remains that if x= 3, the denominator goes poof, and thus undefined
even though the simplified version doesn't show that

Answer 8

wait.. 3 will yield 0/0 there, not a great example

Answer 9

Okay I think I kind of understand. I've just always thought of them as equivalent so this is new to me

Answer 10

trying to think of a better example hehe
but often times the constraints on either a root or an undefined, in the simplified version they get canceled out, and not show as in the original one
but the domain constraints still apply to the original one, thus the simplified version follows the same constraints of the original

Answer 11

okay but they are not equivalent due to the fact that they don't show up in the simplified expression or they are because they are subject to the same domain constraints

Answer 12

$$
y = \cfrac{1}{\sqrt{2x}} \implies y^2 = \cfrac{1}{2x}
$$
maybe a better example
if you notice the original, it has the constraints that if "x" becomes negative
the root turns to "i" or imaginary

notice the simplified version doesn't show that
it only shows a happy 1/2x whose domain is really \(\large (-\infty, +\infty)\)

Answer 13

they'd not be equivalent due to the discrepancies in domains
otherwise, they're equal

Answer 14

ah okay thank you!

Answer 15

yw

Answer 16

Wait! one more question

Answer 17

isn't the example you just showed me irrational as it has a radical in the denominator

Answer 18

well, it could have been \(\large y = \sqrt{\cfrac{1}{2x}} \implies y = \cfrac{\sqrt{1}}{\sqrt{2x}}\)

Answer 19

ah okay so that is a rational expression then

Answer 20

yes, thus it expresses a "ratio" relationshiop, thus "ratio"nal

Answer 21

well, I used 1, but \(1^2 = 1\) so \(\sqrt{1} = \sqrt{1^2} \implies 1\)

Answer 22

but don't the roots still end up becoming imaginary

Answer 23

I'm prty sure even if u simplify like that, you still have to keep restrictions in place

Answer 24

they'd if "x" becomes negative, thus the domain constraint

Answer 25

y^2=1/2(-1) y^2=-0.5 y=sqrt(-0.5)= i sqrt(0.5)

Answer 26

but with domain constraints, im inclined to say yes

Answer 27

but you said that that one resulted in real roots with -infinity or i am misunderstanding which is probably the case

Answer 28

where are u getting infinities olol, should i read up

Answer 29

ya

Answer 30

theres nothing wrong with imaginary roots? im just not seeing the infinity anywhere, unless this is somehow calc

Answer 31

I know theres nothing wrong with them but what i'm trying to say is that when I solve y=1/sqrt(2x) and y^2=1/2x i get the same roots

Answer 32

and i thought i wasn't supposed to get an imaginary root with the second problem

Answer 33

so don't they have the same domain constraints

Answer 34

hehe, no I said, that the simplified version doesn't show the restrictions on the domain for the original, and instead it was showing a domain of \(\large (-\infty, +\infty)\), which would make it non-equivalent to the original

Answer 35

oh okay

Answer 36

so that where the infinities came from hahaha

Answer 37

okay thanks guys