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anonymous
 one year ago
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
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[n _{1}^2 = n _{2}^2\] better formatting

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so taking the square root i should get \[\pm n _{1} = \pm n _{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the correct answer is \[n _{1} = \pm n _{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what was the original problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0proving 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
 one year ago
Best ResponseYou've already chosen the best response.0prove n^2  1 is one to one, onto, or both is the original problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is n positive integers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the problem says the set of all integers for both domain and codomain so positive and negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f(n) = n^2 1 , is not one to one f(3) = f(3) but 3 \( \neq \) 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0\( n_1^2 1 = n_2^2 1\) \( n_1^2 = n_2^2 \) 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
 one year ago
Best ResponseYou've already chosen the best response.0oh, 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
 one year ago
Best ResponseYou've already chosen the best response.0right, squaring undoes the negative (3)^2 = (3)^2 but 3 \( \neq \) 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can do a 'direct' one to one proof with a cubic function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0show that f(n) = n^3 1 is one to one

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1I may be able to help here, use a and b to make your life easier though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also show that your function is not onto either hint use counterexample

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.1If 11 \(f(a)\not=f(b)\) 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 11

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would be one to one for the n^31

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1thinking of a graph helps there. Does it pass the horizontal line test?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1for onto, you show every element has an inverse essentially.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a 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

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1so, let \(a\not=b\) f(a)=f(b)...a=b therefore contradiction

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1try filling in the rest

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1oh and make sure you say a,b elem Z to start

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1forgot that bit...always had a point taken off for it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so y = n^3 1 then y = \[\sqrt[3]{n^3 1}\]

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1are you doing 11 or onto?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was going to do onto

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you cube root a cube, you get \(n _{1}^3 = n _{2}^3 \\ \implies \\ \sqrt[3] {n _{1}^3} = \sqrt[3]{n _{2}^3} \)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1ok, can we just use a,b,c... much less confusing

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1so for onto, I can do the square again

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1so let a be real(in order to be onto, we need this to be a function onto the reals) then a1 is real, and sqrta1 is real and\( f(\sqrt{a1})=(\sqrt{a1})^2+1=a1=+/a\) and +/a is real, therefore there exists an y in R st f(x)=y for every x.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1onto the positive reals* abs. a1 is +R

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1well, to that effect. I messed something up in there

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1X^3+1 would be easier. since no potential complex numbers x^3+1 (x1)^1/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh 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^21) y(1) = sqrt(2) therefore since i get a number not in the original set of integers i could say its not onto?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but i didnt realize that until i saw what you did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if n = 4 which is an integer there is no value such that f(n) = 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i knew it was restricted to integers but i didnt realize i could use that as the reason its not onto

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1yea, pick a number and make it a counter ex

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so for onto there are many integers that f(n) is not in the domain

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1yep, all you need is 1 value that is in the range that has no x to get you there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. 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
 one year ago
Best ResponseYou've already chosen the best response.1... but the x^2 is onto. The range of the function x^2+1 is [1,infty)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1sqrt 4 is 2,+2

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1because it isn't one to one, it can be onto.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1you just run into defining the range being necessary

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1we are defining a positive domain so a sqrt has a real result

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait. maybe i misunderstand onto. but it the problem says the domain and codomain are all integers then y=x^2  1 y=sqrt(x^21) 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
 one year ago
Best ResponseYou've already chosen the best response.0also 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
 one year ago
Best ResponseYou've already chosen the best response.0similarly with even roots same deal. and odd roots

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1you don't have an accurate inverse it's \[y=\sqrt{x1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.1since the squared cancels out it doesn't matter if sqrt(x1) is negative(which it cannot be given the definition of its domain(our prior range)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1you have to do f(y)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1not just compute y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait isnt the inverse y = \[\sqrt{x+1} \] and not y= \[\sqrt{x1}\]

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1f(x)=x^2+1 y=x^2+1 switch x=y^2+1 x1=y^2 sqrt(x1)=y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im 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
 one year ago
Best ResponseYou've already chosen the best response.1to be onto, you must be able to hit every element of the range

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1here our function is Z>[1,infty)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, 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
 one year ago
Best ResponseYou've already chosen the best response.1yea. 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
 one year ago
Best ResponseYou've already chosen the best response.1This slideshow may help http://www.csee.umbc.edu/~stephens/203/PDF/73.pdf

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0every function is onto its range, trivially :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not every function is onto its 'codomain'

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1function implies surjective

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1but you still have to learn to prove it

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1it's part of proving something is a function.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1here 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
 one year ago
Best ResponseYou've already chosen the best response.0ok. slide 7.3.6 is what i need.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im looking at the slides so i might be gone a few minutes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 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
 one year ago
Best ResponseYou've already chosen the best response.1it'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
 one year ago
Best ResponseYou've already chosen the best response.1eg. 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
 one year ago
Best ResponseYou've already chosen the best response.0you call the 'range' what i call the 'codomain'

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1they are the same thing, but we didn't have one defined before

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1ok, maybe not logically the exact same thing :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a function f: N > R , N is domain and R is codomain and f(N) i call range. some people call it the image .

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1I call it image. But to be honest, it's semantics to me. I learned range first it stuck.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0got 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
 one year ago
Best ResponseYou've already chosen the best response.0sorry for the bad formatting but im slow using the equation and inserting them

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1ok, now, leave out the middle bit. Just state a y value

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1the actually finding the inverse is scratch work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i guess 2, 3 or 4 any really

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1oh, no I mean just claim y=inverse essentially

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1you don't show the work there. You do have to show y is in your codomain/range though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so just say if y is in Y the there is an x in Z (the integers) such that f(n) = y

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1that's what you are proving.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i thought i just proved it.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1"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
 one year ago
Best ResponseYou've already chosen the best response.1just a restatement needed.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. just for got 2 of the needed clarification statements. i see why i need them.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1Alright, so no promises on this being perfect, but I hope I helped

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was excited and typing fast because the scratch work was where i was stuck

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1I'm gonna try and nap an hour or two before my test

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you and jayzzd i really appreciate it. i like those slides.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1so, good morning/night and np

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0test!! dang i would be asleep also

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1praxis II math :( :( hence the not sleeping now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i see well good luck.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0example: f = { (1, 4 ) , ( 2 , 2) ( 3 , 7 ) }  f  =  { 1,2,3} 

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.1thanks, you too night all
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