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\[n _{1}^2 = n _{2}^2\]
better formatting

so taking the square root i should get \[\pm n _{1} = \pm n _{2}\]

but the correct answer is \[n _{1} = \pm n _{2}\]

what was the original problem

prove n^2 - 1 is one to one, onto, or both is the original problem

is n positive integers?

the problem says the set of all integers for both domain and codomain
so positive and negative

f(n) = n^2 -1 , is not one to one
f(3) = f(-3)
but 3 \( \neq \) -3

right, squaring undoes the negative
(3)^2 = (-3)^2
but 3 \( \neq \) -3

you can do a 'direct' one to one proof with a cubic function

show that f(n) = n^3 -1 is one to one

I may be able to help here, use a and b to make your life easier though

also show that your function is not onto either
hint use counterexample

so it would be one to one for the n^3-1

thinking of a graph helps there. Does it pass the horizontal line test?

for onto,
you show every element has an inverse essentially.

a graph does help i knw in my head that x^3 would be one to one so i just had to to a small proof

yes it does pass

so, let \(a\not=b\)
f(a)=f(b)...a=b therefore contradiction

try filling in the rest

oh and make sure you say a,b elem Z to start

forgot that bit...always had a point taken off for it

ok so y = n^3 -1
then y = \[\sqrt[3]{n^3 -1}\]

wait,

are you doing 1-1 or onto?

i was going to do onto

\( n_1 = n_2\)

ok, can we just use a,b,c... much less confusing

so for onto, I can do the square again

onto the positive reals* abs. a-1 is +R

well, to that effect. I messed something up in there

ok

X^3+1 would be easier. since no potential complex numbers x^3+1
(x-1)^1/3

but i didnt realize that until i saw what you did

if n = -4 which is an integer
there is no value such that
f(n) = -4

i knew it was restricted to integers but i didnt realize i could use that as the reason its not onto

yea, pick a number and make it a counter ex

ok so for onto there are many integers that f(n) is not in the domain

yep, all you need is 1 value that is in the range that has no x to get you there

... but the x^2 is onto. The range of the function x^2+1 is [1,infty)

sqrt 4 is 2,+-2

because it isn't one to one, it can be onto.

you just run into defining the range being necessary

we are defining a positive domain so a sqrt has a real result

similarly with even roots same deal.
and odd roots

you don't have an accurate inverse it's \[y=\sqrt{x-1}\]

you have to do f(y)

not just compute y

wait isnt the inverse y = \[\sqrt{x+1} \] and not y= \[\sqrt{x-1}\]

f(x)=x^2+1
y=x^2+1
switch
x=y^2+1
x-1=y^2
sqrt(x-1)=y

to be onto, you must be able to hit every element of the range

here our function is Z->[1,infty)

This slideshow may help http://www.csee.umbc.edu/~stephens/203/PDF/7-3.pdf

every function is onto its range, trivially :)

not every function is onto its 'codomain'

function implies surjective

but you still have to learn to prove it

it's part of proving something is a function.

ok. slide 7.3.6 is what i need.

im looking at the slides so i might be gone a few minutes

is cool

you call the 'range' what i call the 'codomain'

they are the same thing, but we didn't have one defined before

ok, maybe not logically the exact same thing :P

I call it image. But to be honest, it's semantics to me. I learned range first it stuck.

sorry for the bad formatting but im slow using the equation and inserting them

ok, now, leave out the middle bit. Just state a y value

the actually finding the inverse is scratch work

i guess 2, 3 or 4 any really

oh well dang.

oh, no I mean just claim y=inverse essentially

you don't show the work there. You do have to show y is in your codomain/range though

so just say if y is in Y the there is an x in Z (the integers) such that f(n) = y

that's what you are proving.

i thought i just proved it.

just a restatement needed.

ok. just for got 2 of the needed clarification statements. i see why i need them.

Alright, so no promises on this being perfect, but I hope I helped

i was excited and typing fast because the scratch work was where i was stuck

you did

I'm gonna try and nap an hour or two before my test

you and jayzzd i really appreciate it. i like those slides.

so, good morning/night and np

test!! dang i would be asleep also

praxis II math :( :( hence the not sleeping now

i see well good luck.

example:
f = { (1, 4 ) , ( 2 , 2) ( 3 , 7 ) }
| f | = | { 1,2,3} |

thanks, you too night all