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

1. anonymous

$n _{1}^2 = n _{2}^2$ better formatting

2. anonymous

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

3. anonymous

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

4. anonymous

what was the original problem

5. 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

6. anonymous

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

7. anonymous

is n positive integers?

8. anonymous

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

9. anonymous

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

10. 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

11. 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

12. 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

13. 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 $$n_1 \neq n_2$$

14. 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

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. FibonacciChick666

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

19. anonymous

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

20. 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

21. 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

22. anonymous

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

23. FibonacciChick666

yep

24. FibonacciChick666

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

25. FibonacciChick666

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

26. 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

27. anonymous

yes it does pass

28. 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

29. FibonacciChick666

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

30. FibonacciChick666

try filling in the rest

31. FibonacciChick666

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

32. FibonacciChick666

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

33. anonymous

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

34. FibonacciChick666

wait,

35. FibonacciChick666

are you doing 1-1 or onto?

36. anonymous

i was going to do onto

37. anonymous

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

38. anonymous

$$n_1 = n_2$$

39. FibonacciChick666

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

40. FibonacciChick666

so for onto, I can do the square again

41. 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.

42. FibonacciChick666

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

43. FibonacciChick666

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

44. anonymous

ok

45. FibonacciChick666

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

46. 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?

47. anonymous

but i didnt realize that until i saw what you did

48. anonymous

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

49. anonymous

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

50. FibonacciChick666

yea, pick a number and make it a counter ex

51. anonymous

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

52. FibonacciChick666

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

53. 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.

54. FibonacciChick666

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

55. FibonacciChick666

sqrt 4 is 2,+-2

56. FibonacciChick666

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

57. FibonacciChick666

you just run into defining the range being necessary

58. FibonacciChick666

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

59. 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?

60. 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 $$\large n_2 = n_1$$

61. anonymous

similarly with even roots same deal. and odd roots

62. FibonacciChick666

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

63. 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?

64. 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)

65. FibonacciChick666

you have to do f(y)

66. FibonacciChick666

not just compute y

67. anonymous

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

68. FibonacciChick666

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

69. 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

70. FibonacciChick666

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

71. FibonacciChick666

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

72. 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

73. 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.

74. FibonacciChick666

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

75. anonymous

every function is onto its range, trivially :)

76. anonymous

not every function is onto its 'codomain'

77. FibonacciChick666

function implies surjective

78. FibonacciChick666

but you still have to learn to prove it

79. FibonacciChick666

it's part of proving something is a function.

80. 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

81. anonymous

ok. slide 7.3.6 is what i need.

82. anonymous

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

83. FibonacciChick666

is cool

84. 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.

85. 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

86. 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.

87. anonymous

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

88. FibonacciChick666

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

89. FibonacciChick666

ok, maybe not logically the exact same thing :P

90. 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 .

91. FibonacciChick666

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

92. 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

93. anonymous

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

94. FibonacciChick666

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

95. FibonacciChick666

the actually finding the inverse is scratch work

96. anonymous

i guess 2, 3 or 4 any really

97. anonymous

oh well dang.

98. FibonacciChick666

oh, no I mean just claim y=inverse essentially

99. FibonacciChick666

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

100. anonymous

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

101. FibonacciChick666

that's what you are proving.

102. anonymous

i thought i just proved it.

103. 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 $$\color\red{\text{if y is in Y then there is an x in Z (the integers) such that f(n) = y}}$$ "

104. FibonacciChick666

just a restatement needed.

105. anonymous

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

106. FibonacciChick666

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

107. anonymous

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

108. anonymous

you did

109. FibonacciChick666

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

110. anonymous

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

111. FibonacciChick666

so, good morning/night and np

112. anonymous

test!! dang i would be asleep also

113. FibonacciChick666

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

114. anonymous

i see well good luck.

115. 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.

116. anonymous

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

117. FibonacciChick666

thanks, you too night all