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

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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

Algebra
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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 \[n _{1} = \pm n _{2}\]

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what was the original problem
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
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
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
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
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
\( n_1^2 -1 = n_2^2 -1\) \( n_1^2 = n_2^2 \) 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 \( n_1 \neq n_2 \)
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
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
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
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
so it would be one to one for the n^3-1
yep
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 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
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
if you cube root a cube, you get \(n _{1}^3 = n _{2}^3 \\ \implies \\ \sqrt[3] {n _{1}^3} = \sqrt[3]{n _{2}^3} \)
\( n_1 = n_2\)
ok, can we just use a,b,c... much less confusing
so for onto, I can do the square again
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.
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
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?
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
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.
... 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
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?
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 \( \large n_2 = n_1 \)
similarly with even roots same deal. and odd roots
you don't have an accurate inverse it's \[y=\sqrt{x-1}\]
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?
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)
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
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
to be onto, you must be able to hit every element of the range
here our function is Z->[1,infty)
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
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.
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.
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
ok. slide 7.3.6 is what i need.
im looking at the slides so i might be gone a few minutes
is cool
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.
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
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.
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
a function f: N -> R , N is domain and R is codomain and f(N) i call range. some people call it the image .
I call it image. But to be honest, it's semantics to me. I learned range first it stuck.
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
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.
"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 \(\color\red{\text{if y is in Y then there is an x in Z (the integers) such that f(n) = y}}\) "
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.
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.
example: f = { (1, 4 ) , ( 2 , 2) ( 3 , 7 ) } | f | = | { 1,2,3} |
thanks, you too night all

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