#inequality
48 messages · Page 1 of 1 (latest)
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Does $x^2 + 1 = x(x + 1)$?
Techie Literate
no
Right.
tysm
(You can also just multiply both sides by x^2+1 from the start)
Since it is not being compared to 0
Then a trivial case will pop up
so as to cancel x^2+1 in the denominator on LHS?
Yes.
?
TY!
Yep.
also, a small ques, are x E R and x E (-inf, -1] u (0, inf) the same?
No, it excludes $x\in (-1, 0)$.
Kocher
∈=E(just copied frm google)
The latter, at least.
but
$x\in\mathbb{R}\implies x\in (-\infty, \infty)$
Kocher
Is this set you proposed:
- Continuous?
- Contain the reals on the discontinuous intervals?
The answer to both these question is no, so it isn’t the reals.
set as in the wrong set i proposed?
Yes.
Only one of these questions needs to be false in order for it to not be the set of reals.
ok, i'll keep these in my mind
hey
- what does ''Contain the reals on the discontinuous intervals'' mean tho
if its discontinuous set then shoulnt it already 'not be real'/?
- and apart from R and (-inf,inf) there's no way of representing a set containing all real numbers?
Yeah, I was just adding a little extra.
rationalUirrational
r u answering the 1st ques?
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