square root question
i am at the end of a problem and i have n1^2 = n2^2 so taking the square root shouldn't i get +or- n1 = = or - n2

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

what was the original problem

- anonymous

proving a function is one to one. its a discrete math problem but for some reason i seem to have forgot how to do algebra

- anonymous

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

- anonymous

is n positive integers?

- anonymous

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

- anonymous

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

- anonymous

thats what i thought but was trying to prove it abstractly. but i guess i could just show one counter example and that would be proof enough it is not one to one

- anonymous

so the problem to prove it directly
\( n_1^2 -1 = n_2^2 -1 \)
\( n_1^2 = n_2^2 \)
you can't get n_1 = n_2 from this

- anonymous

the our teacher did it was to assume that it is one to one and try and prove it or prove otherwise. We basically take the equation and set it equal to itself with the variable term being n1, n2 etc. which was weird to me

- anonymous

\( n_1^2 -1 = n_2^2 -1\)
\( n_1^2 = n_2^2 \)
you have to stop here. you cannot deduce that \( n_1 = n_2 \)
In general any counterexample works as follows:
\(n_1 = -n_2 \) and \( n_1 \neq 0 \)
then \( n_1^2 = (-n_2)^2 = n_2^2 \)
but \( n_1 \neq n_2 \)

- anonymous

oh, i see. i was not getting the algebra step you did. n1 = -n2 because its square they are equal and have multiple "x" values mapping to y values

- anonymous

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

- anonymous

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

- anonymous

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

- FibonacciChick666

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

- anonymous

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

- anonymous

ok so \[n _{1}^3 = n _{2}^3\]
then cube root them
n1 = -n2
so for example
f(3) and f(-3)
26 does not = -26

- FibonacciChick666

If 1-1 \(f(a)\not=f(b)\)
so for the squared one, let a be an integer
\[f(a)=a^2+1\]
\[f(-a)=(-a)^2+1=a^2+1\]
Therefore f(a)=f(-a) and f is not 1-1

- anonymous

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

- FibonacciChick666

yep

- FibonacciChick666

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

- FibonacciChick666

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

- anonymous

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

- anonymous

yes it does pass

- FibonacciChick666

so you use f(a) and f(b) for the n^3 one, just like I did for n^2 but in this instance you will come to the conclusion that they only equal if a=b which you stipulate is not possible at the beginning

- FibonacciChick666

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

- FibonacciChick666

try filling in the rest

- FibonacciChick666

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

- FibonacciChick666

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

- anonymous

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

- FibonacciChick666

wait,

- FibonacciChick666

are you doing 1-1 or onto?

- anonymous

i was going to do onto

- anonymous

if you cube root a cube, you get
\(n _{1}^3 = n _{2}^3
\\ \implies
\\ \sqrt[3] {n _{1}^3} = \sqrt[3]{n _{2}^3}
\)

- anonymous

\( n_1 = n_2\)

- FibonacciChick666

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

- FibonacciChick666

so for onto, I can do the square again

- FibonacciChick666

so let a be real(in order to be onto, we need this to be a function onto the reals)
then |a-1| is real, and sqrt|a-1| is real
and\( f(\sqrt{|a-1|})=(\sqrt{|a-1|})^2+1=a-1=+/-a\) and +/-a is real, therefore there exists an y in R st f(x)=y for every x.

- FibonacciChick666

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

- FibonacciChick666

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

- anonymous

ok

- FibonacciChick666

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

- anonymous

oh for the n^2 -1 the problem was restricted to integers. So for onto could i just do
y=x^2 - 1
y=sqrt(x^2-1)
y(1) = sqrt(2)
therefore since i get a number not in the original set of integers i could say its not onto?

- anonymous

but i didnt realize that until i saw what you did

- anonymous

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

- anonymous

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

- FibonacciChick666

yea, pick a number and make it a counter ex

- anonymous

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

- FibonacciChick666

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

- anonymous

ok. I think i get it now. its easy to show its not one to one and not onto. just find a counter example. proving it is one to one or onto is going to be a little harder.

- FibonacciChick666

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

- FibonacciChick666

sqrt 4 is 2,+-2

- FibonacciChick666

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

- FibonacciChick666

you just run into defining the range being necessary

- FibonacciChick666

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

- anonymous

wait. maybe i misunderstand onto. but it the problem says the domain and codomain are all integers then
y=x^2 - 1
y=sqrt(x^2-1)
y(3) = sqrt(3^2 -1)
y=(sqrt(8))
therefore since i get a number not in the integers wouldnt it be not onto?

- anonymous

also it might help remember
for \( \large x \in \mathbb R \) , \( \large \sqrt{x^2}= |x| \)
but \( \large \sqrt[3]{x^3}= x \)
Using this
if \(\large n_2^2 = n_1^2 \)
if we square root both sides
we get
\(\large \sqrt{n_2^2 } = \sqrt{n_1^2} \iff |n_2| = |n_1| \)
but you cannot remove the absolute value bars to get \( \large n_2 = n_1 \)

- anonymous

similarly with even roots same deal.
and odd roots

- FibonacciChick666

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

- anonymous

oh ok. but if x is 8 then i get sqrt(7) which is still not an integer. so its not it the set of integers specified by the problem?

- FibonacciChick666

since the squared cancels out it doesn't matter if sqrt(x-1) is negative(which it cannot be given the definition of its domain(our prior range)

- FibonacciChick666

you have to do f(y)

- FibonacciChick666

not just compute y

- anonymous

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

- FibonacciChick666

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

- anonymous

im still not seeing how it is onto. i say its not because there are integers the domain that make the codomain not in the set of integers

- FibonacciChick666

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

- FibonacciChick666

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

- anonymous

oh, ok. so even though there are numbers not in the range. there are numbers that can be found for all integers which is our range

- FibonacciChick666

yea. That function is not onto the integers. It is onto a subset of the integers. So we are not going backwards from the whole integers, that wasn't our range. We are just taking our range and going back.

- FibonacciChick666

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

- anonymous

every function is onto its range, trivially :)

- anonymous

not every function is onto its 'codomain'

- FibonacciChick666

function implies surjective

- FibonacciChick666

but you still have to learn to prove it

- FibonacciChick666

it's part of proving something is a function.

- FibonacciChick666

here is the thing though, if a problem were to say f(x):R-->R then it would have to span the whole reals. we weren't given something like that here, so we assume the range makes sense

- anonymous

ok. slide 7.3.6 is what i need.

- anonymous

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

- FibonacciChick666

is cool

- anonymous

but i really appreciate both of you guys helping. the one to one isnt that bad. the onto i think is where i need practice.

- FibonacciChick666

it's not as terrible as it feels. Do a checklist: what is my range? Do I hit my whole range? yes?-->onto no?--->not onto

- FibonacciChick666

eg. f(x):Z-->Z
f(x)=x^2+4
What's my range? Z
Do I hit every point? No
Not onto. e.g. Let y=0
x^2=-4-->x=2i which is not in Z.

- anonymous

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

- FibonacciChick666

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

- FibonacciChick666

ok, maybe not logically the exact same thing :P

- anonymous

a function
f: N -> R , N is domain and R is codomain
and f(N) i call range. some people call it the image .

- FibonacciChick666

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

- anonymous

got it. Im going to use the letters in my problem
f(n) = n^2 -1
Let y be in Y
n^2 - 1 = y
n^2 = y+1
n= sqrt(y +1)
f(n) = f(sqrt(y+1)) = (sqrt (y+1))^2 -1 = y +1 -1 = y
therefore it exists

- anonymous

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

- FibonacciChick666

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

- FibonacciChick666

the actually finding the inverse is scratch work

- anonymous

i guess 2, 3 or 4 any really

- anonymous

oh well dang.

- FibonacciChick666

oh, no I mean just claim y=inverse essentially

- FibonacciChick666

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

- anonymous

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

- FibonacciChick666

that's what you are proving.

- anonymous

i thought i just proved it.

- FibonacciChick666

"f(n) = n^2 -1
Let y be in Y
\(\color\red{\text{y+1 is in Z and sqrt(y+1) is in Z}}\)
f(n) = f(sqrt(y+1)) = (sqrt (y+1))^2 -1 = y +1 -1 = y
therefore \(\color\red{\text{if y is in Y then there is an x in Z (the integers) such that f(n) = y}}\)
"

- FibonacciChick666

just a restatement needed.

- anonymous

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

- FibonacciChick666

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

- anonymous

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

- anonymous

you did

- FibonacciChick666

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

- anonymous

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

- FibonacciChick666

so, good morning/night and np

- anonymous

test!! dang i would be asleep also

- FibonacciChick666

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

- anonymous

i see well good luck.

- anonymous

I like the slides too. interesting | f | = | X | I am assuming this is because f is the set of all ordered pairs , and the size of this is determined by the x values . each x value has a y output.

- anonymous

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

- FibonacciChick666

thanks, you too night all

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