Question

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

Answer 1

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

better formatting

Answer 2

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

Answer 3

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

Answer 4

what was the original problem

Answer 5

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

Answer 6

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

Answer 7

is n positive integers?

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

\( n_1^2 -1 = n_2^2 -1\)
\( n_1^2 = n_2^2 \)

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 \( n_1 \neq n_2 \)

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

yep

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

yes it does pass

Answer 28

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

Answer 29

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

Answer 30

try filling in the rest

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

wait,

Answer 35

are you doing 1-1 or onto?

Answer 36

i was going to do onto

Answer 37

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}

\)

Answer 38

\( n_1 = n_2\)

Answer 39

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

Answer 40

so for onto, I can do the square again

Answer 41

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.

Answer 42

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

Answer 43

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

Answer 44

ok

Answer 45

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

Answer 46

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?

Answer 47

but i didnt realize that until i saw what you did

Answer 48

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

Answer 49

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

Answer 50

yea, pick a number and make it a counter ex

Answer 51

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

Answer 52

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

Answer 53

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.

Answer 54

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

Answer 55

sqrt 4 is 2,+-2

Answer 56

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

Answer 57

you just run into defining the range being necessary

Answer 58

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

Answer 59

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?

Answer 60

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 \( \large n_2 = n_1 \)

Answer 61

similarly with even roots same deal.
and odd roots

Answer 62

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

Answer 63

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?

Answer 64

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)

Answer 65

you have to do f(y)

Answer 66

not just compute y

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

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.

Answer 74

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

Answer 75

every function is onto its range, trivially :)

Answer 76

not every function is onto its 'codomain'

Answer 77

function implies surjective

Answer 78

but you still have to learn to prove it

Answer 79

it's part of proving something is a function.

Answer 80

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

Answer 81

ok. slide 7.3.6 is what i need.

Answer 82

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

Answer 83

is cool

Answer 84

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.

Answer 85

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

Answer 86

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.

Answer 87

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

Answer 88

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

Answer 89

ok, maybe not logically the exact same thing :P

Answer 90

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

Answer 91

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

Answer 92

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

Answer 93

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

Answer 94

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

Answer 95

the actually finding the inverse is scratch work

Answer 96

i guess 2, 3 or 4 any really

Answer 97

oh well dang.

Answer 98

oh, no I mean just claim y=inverse essentially