set theorem basic exercise

[elder cave]

I don’t understand. I want to know (i) and then I can try the rest on my own.

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[candid notch]

I guess just draw the region

i says it's all real numbers that are negative.. so draw that

[elder cave]

Just on a line?

[candid notch]

draw the entire line.. then shade in the region

and that's wrong regardless

[elder cave]

but that’s like infinite wdym

[elder cave]

Yeah cuz it’s just all negative integers

but I’m confused

idk how the answer is supposed to look like

[candid notch]

no...

unless you cropped out the instructions that says the universal set is Z

draw the real line

then shade in all negative numbers

hint: arrows exist to show the line continues on

[elder cave]

uh

what is the real line I don’t get it

Literally all of the before was just set theory proofs

And then I get this random exercise

[candid notch]

the real numbers

... as a line

https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(Ellis_and_Burzynski)/02%3A_Basic_Properties_of_Real_Numbers/2.03%3A_The_Real_Number_Line_and_the_Real_Numbers

[elder cave]

Oh yea

so like

[candid notch]

yes, though draw a line not a line segment

This is a line, shade in the negatives

[elder cave]

Then (ii)

{ x | |x| < 1 }

the absolute value of x is smaller than 1

so it’s just everything left to 1

?

[manic notch]

Nope

-1 < x < 1

[candid notch]

I mean, add the ) at 0 to make it crystal..

but yes, as Ravi barged in and said... |x|<a iff -a<x<a

[elder cave]

crystal?

[candid notch]

crystal clear.......

[elder cave]

you’re speaking way too vague and annoying

[candid notch]

if you want clarification, ask specific questions

[elder cave]

Okay so the |x| in this case is not the same as absolute value?

[candid notch]

it is

[elder cave]

or is it

[candid notch]

|x| is the abs value of x

[manic notch]

Ouch

[elder cave]

Yeah however I understood the function as like if you put any number it’s just a positive output

[candid notch]

it is

or 0

[elder cave]

so how did we get to -1<x<1

I get the x<1 part

where’d the -1 come from

[candid notch]

if $x\in(-1,0)$, then $\abs{x}=-x\in(0,1)$

[native drift]

Omegabet_

[candid notch]

since x is negative

so it's clear $(-1,1)\subseteq{x\mid\abs{x}<1}$

[native drift]

Omegabet_

[candid notch]

and it's clear that anything with magnitude less than 1 cant go outside of (-1,1)

it all follows from the usual piecewise definition

[elder cave]

Yeah I understand

It can’t be -1 or lower cuz then it’s >=1

which is the opposite of <1

Okay (iii)

now uh idk what the epsilon signifies

[candid notch]

it's a number

just use the property we just discussed.

[elder cave]

-ε< x-a < ε

[candid notch]

yes

so $x\in(a-\varepsilon,a+\varepsilon)$

[native drift]

Omegabet_

[candid notch]

ie an (open) interval around a of radius eps

[elder cave]

could you elaborate a bit more

[candid notch]

no

not really

(a-e,a+e) is an interval, open, centered at a

and has radius epsilon

[elder cave]

I understand the radius part being centered at a

I don’t get the interval part tho

isn’t that just saying it’s a radius epsilon centered at a

can we assume some random numbers for epsilon and a

a = 2 and epsilon = 5 e.g

so I can get a better picture of how this works

[candid notch]

it's an interval...

(a,b) is an interval

(a-e,a+e) is an interval

$(a,b):={x\mid a<x<b}$

[native drift]

Omegabet_

[candid notch]

$[a,b]:={x\mid a\leq x\leq b}$

[native drift]

Omegabet_

[candid notch]

where the universal set is R ofc

[elder cave]

Oh yeah okay I am just not so familiar with English terms das all