anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[n _{1}^2 = n _{2}^2\] better formatting
anonymous
  • anonymous
so taking the square root i should get \[\pm n _{1} = \pm n _{2}\]
anonymous
  • anonymous
but the correct answer is \[n _{1} = \pm n _{2}\]

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anonymous
  • anonymous
what was the original problem
anonymous
  • 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
  • anonymous
prove n^2 - 1 is one to one, onto, or both is the original problem
anonymous
  • anonymous
is n positive integers?
anonymous
  • anonymous
the problem says the set of all integers for both domain and codomain so positive and negative
anonymous
  • anonymous
f(n) = n^2 -1 , is not one to one f(3) = f(-3) but 3 \( \neq \) -3
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
right, squaring undoes the negative (3)^2 = (-3)^2 but 3 \( \neq \) -3
anonymous
  • anonymous
you can do a 'direct' one to one proof with a cubic function
anonymous
  • anonymous
show that f(n) = n^3 -1 is one to one
FibonacciChick666
  • FibonacciChick666
I may be able to help here, use a and b to make your life easier though
anonymous
  • anonymous
also show that your function is not onto either hint use counterexample
anonymous
  • 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
  • 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
  • anonymous
so it would be one to one for the n^3-1
FibonacciChick666
  • FibonacciChick666
yep
FibonacciChick666
  • FibonacciChick666
thinking of a graph helps there. Does it pass the horizontal line test?
FibonacciChick666
  • FibonacciChick666
for onto, you show every element has an inverse essentially.
anonymous
  • 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
  • anonymous
yes it does pass
FibonacciChick666
  • 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
  • FibonacciChick666
so, let \(a\not=b\) f(a)=f(b)...a=b therefore contradiction
FibonacciChick666
  • FibonacciChick666
try filling in the rest
FibonacciChick666
  • FibonacciChick666
oh and make sure you say a,b elem Z to start
FibonacciChick666
  • FibonacciChick666
forgot that bit...always had a point taken off for it
anonymous
  • anonymous
ok so y = n^3 -1 then y = \[\sqrt[3]{n^3 -1}\]
FibonacciChick666
  • FibonacciChick666
wait,
FibonacciChick666
  • FibonacciChick666
are you doing 1-1 or onto?
anonymous
  • anonymous
i was going to do onto
anonymous
  • 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
  • anonymous
\( n_1 = n_2\)
FibonacciChick666
  • FibonacciChick666
ok, can we just use a,b,c... much less confusing
FibonacciChick666
  • FibonacciChick666
so for onto, I can do the square again
FibonacciChick666
  • 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
  • FibonacciChick666
onto the positive reals* abs. a-1 is +R
FibonacciChick666
  • FibonacciChick666
well, to that effect. I messed something up in there
anonymous
  • anonymous
ok
FibonacciChick666
  • FibonacciChick666
X^3+1 would be easier. since no potential complex numbers x^3+1 (x-1)^1/3
anonymous
  • 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
  • anonymous
but i didnt realize that until i saw what you did
anonymous
  • anonymous
if n = -4 which is an integer there is no value such that f(n) = -4
anonymous
  • anonymous
i knew it was restricted to integers but i didnt realize i could use that as the reason its not onto
FibonacciChick666
  • FibonacciChick666
yea, pick a number and make it a counter ex
anonymous
  • anonymous
ok so for onto there are many integers that f(n) is not in the domain
FibonacciChick666
  • FibonacciChick666
yep, all you need is 1 value that is in the range that has no x to get you there
anonymous
  • 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
  • FibonacciChick666
... but the x^2 is onto. The range of the function x^2+1 is [1,infty)
FibonacciChick666
  • FibonacciChick666
sqrt 4 is 2,+-2
FibonacciChick666
  • FibonacciChick666
because it isn't one to one, it can be onto.
FibonacciChick666
  • FibonacciChick666
you just run into defining the range being necessary
FibonacciChick666
  • FibonacciChick666
we are defining a positive domain so a sqrt has a real result
anonymous
  • 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
  • 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
  • anonymous
similarly with even roots same deal. and odd roots
FibonacciChick666
  • FibonacciChick666
you don't have an accurate inverse it's \[y=\sqrt{x-1}\]
anonymous
  • 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
  • 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
  • FibonacciChick666
you have to do f(y)
FibonacciChick666
  • FibonacciChick666
not just compute y
anonymous
  • anonymous
wait isnt the inverse y = \[\sqrt{x+1} \] and not y= \[\sqrt{x-1}\]
FibonacciChick666
  • FibonacciChick666
f(x)=x^2+1 y=x^2+1 switch x=y^2+1 x-1=y^2 sqrt(x-1)=y
anonymous
  • 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
  • FibonacciChick666
to be onto, you must be able to hit every element of the range
FibonacciChick666
  • FibonacciChick666
here our function is Z->[1,infty)
anonymous
  • 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
  • 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
  • FibonacciChick666
This slideshow may help http://www.csee.umbc.edu/~stephens/203/PDF/7-3.pdf
anonymous
  • anonymous
every function is onto its range, trivially :)
anonymous
  • anonymous
not every function is onto its 'codomain'
FibonacciChick666
  • FibonacciChick666
function implies surjective
FibonacciChick666
  • FibonacciChick666
but you still have to learn to prove it
FibonacciChick666
  • FibonacciChick666
it's part of proving something is a function.
FibonacciChick666
  • 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
  • anonymous
ok. slide 7.3.6 is what i need.
anonymous
  • anonymous
im looking at the slides so i might be gone a few minutes
FibonacciChick666
  • FibonacciChick666
is cool
anonymous
  • 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
  • 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
  • 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
  • anonymous
you call the 'range' what i call the 'codomain'
FibonacciChick666
  • FibonacciChick666
they are the same thing, but we didn't have one defined before
FibonacciChick666
  • FibonacciChick666
ok, maybe not logically the exact same thing :P
anonymous
  • 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
  • FibonacciChick666
I call it image. But to be honest, it's semantics to me. I learned range first it stuck.
anonymous
  • 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
  • anonymous
sorry for the bad formatting but im slow using the equation and inserting them
FibonacciChick666
  • FibonacciChick666
ok, now, leave out the middle bit. Just state a y value
FibonacciChick666
  • FibonacciChick666
the actually finding the inverse is scratch work
anonymous
  • anonymous
i guess 2, 3 or 4 any really
anonymous
  • anonymous
oh well dang.
FibonacciChick666
  • FibonacciChick666
oh, no I mean just claim y=inverse essentially
FibonacciChick666
  • FibonacciChick666
you don't show the work there. You do have to show y is in your codomain/range though
anonymous
  • anonymous
so just say if y is in Y the there is an x in Z (the integers) such that f(n) = y
FibonacciChick666
  • FibonacciChick666
that's what you are proving.
anonymous
  • anonymous
i thought i just proved it.
FibonacciChick666
  • 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
  • FibonacciChick666
just a restatement needed.
anonymous
  • anonymous
ok. just for got 2 of the needed clarification statements. i see why i need them.
FibonacciChick666
  • FibonacciChick666
Alright, so no promises on this being perfect, but I hope I helped
anonymous
  • anonymous
i was excited and typing fast because the scratch work was where i was stuck
anonymous
  • anonymous
you did
FibonacciChick666
  • FibonacciChick666
I'm gonna try and nap an hour or two before my test
anonymous
  • anonymous
you and jayzzd i really appreciate it. i like those slides.
FibonacciChick666
  • FibonacciChick666
so, good morning/night and np
anonymous
  • anonymous
test!! dang i would be asleep also
FibonacciChick666
  • FibonacciChick666
praxis II math :( :( hence the not sleeping now
anonymous
  • anonymous
i see well good luck.
anonymous
  • 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
  • anonymous
example: f = { (1, 4 ) , ( 2 , 2) ( 3 , 7 ) } | f | = | { 1,2,3} |
FibonacciChick666
  • FibonacciChick666
thanks, you too night all

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