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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

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  1. anonymous
    • one year ago
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    \[n _{1}^2 = n _{2}^2\] better formatting

  2. anonymous
    • one year ago
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    so taking the square root i should get \[\pm n _{1} = \pm n _{2}\]

  3. anonymous
    • one year ago
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    but the correct answer is \[n _{1} = \pm n _{2}\]

  4. anonymous
    • one year ago
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    what was the original problem

  5. anonymous
    • one year ago
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    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
    • one year ago
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    prove n^2 - 1 is one to one, onto, or both is the original problem

  7. anonymous
    • one year ago
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    is n positive integers?

  8. anonymous
    • one year ago
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    the problem says the set of all integers for both domain and codomain so positive and negative

  9. anonymous
    • one year ago
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    f(n) = n^2 -1 , is not one to one f(3) = f(-3) but 3 \( \neq \) -3

  10. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \( n_1^2 -1 = n_2^2 -1\) \( n_1^2 = n_2^2 \) 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
    • one year ago
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    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
    • one year ago
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    right, squaring undoes the negative (3)^2 = (-3)^2 but 3 \( \neq \) -3

  16. anonymous
    • one year ago
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    you can do a 'direct' one to one proof with a cubic function

  17. anonymous
    • one year ago
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    show that f(n) = n^3 -1 is one to one

  18. FibonacciChick666
    • one year ago
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    I may be able to help here, use a and b to make your life easier though

  19. anonymous
    • one year ago
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    also show that your function is not onto either hint use counterexample

  20. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    so it would be one to one for the n^3-1

  23. FibonacciChick666
    • one year ago
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    yep

  24. FibonacciChick666
    • one year ago
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    thinking of a graph helps there. Does it pass the horizontal line test?

  25. FibonacciChick666
    • one year ago
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    for onto, you show every element has an inverse essentially.

  26. anonymous
    • one year ago
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    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
    • one year ago
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    yes it does pass

  28. FibonacciChick666
    • one year ago
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    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
    • one year ago
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    so, let \(a\not=b\) f(a)=f(b)...a=b therefore contradiction

  30. FibonacciChick666
    • one year ago
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    try filling in the rest

  31. FibonacciChick666
    • one year ago
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    oh and make sure you say a,b elem Z to start

  32. FibonacciChick666
    • one year ago
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    forgot that bit...always had a point taken off for it

  33. anonymous
    • one year ago
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    ok so y = n^3 -1 then y = \[\sqrt[3]{n^3 -1}\]

  34. FibonacciChick666
    • one year ago
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    wait,

  35. FibonacciChick666
    • one year ago
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    are you doing 1-1 or onto?

  36. anonymous
    • one year ago
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    i was going to do onto

  37. anonymous
    • one year ago
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    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
    • one year ago
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    \( n_1 = n_2\)

  39. FibonacciChick666
    • one year ago
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    ok, can we just use a,b,c... much less confusing

  40. FibonacciChick666
    • one year ago
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    so for onto, I can do the square again

  41. FibonacciChick666
    • one year ago
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    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
    • one year ago
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    onto the positive reals* abs. a-1 is +R

  43. FibonacciChick666
    • one year ago
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    well, to that effect. I messed something up in there

  44. anonymous
    • one year ago
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    ok

  45. FibonacciChick666
    • one year ago
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    X^3+1 would be easier. since no potential complex numbers x^3+1 (x-1)^1/3

  46. anonymous
    • one year ago
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    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
    • one year ago
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    but i didnt realize that until i saw what you did

  48. anonymous
    • one year ago
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    if n = -4 which is an integer there is no value such that f(n) = -4

  49. anonymous
    • one year ago
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    i knew it was restricted to integers but i didnt realize i could use that as the reason its not onto

  50. FibonacciChick666
    • one year ago
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    yea, pick a number and make it a counter ex

  51. anonymous
    • one year ago
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    ok so for onto there are many integers that f(n) is not in the domain

  52. FibonacciChick666
    • one year ago
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    yep, all you need is 1 value that is in the range that has no x to get you there

  53. anonymous
    • one year ago
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    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
    • one year ago
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    ... but the x^2 is onto. The range of the function x^2+1 is [1,infty)

  55. FibonacciChick666
    • one year ago
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    sqrt 4 is 2,+-2

  56. FibonacciChick666
    • one year ago
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    because it isn't one to one, it can be onto.

  57. FibonacciChick666
    • one year ago
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    you just run into defining the range being necessary

  58. FibonacciChick666
    • one year ago
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    we are defining a positive domain so a sqrt has a real result

  59. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    similarly with even roots same deal. and odd roots

  62. FibonacciChick666
    • one year ago
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    you don't have an accurate inverse it's \[y=\sqrt{x-1}\]

  63. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    you have to do f(y)

  66. FibonacciChick666
    • one year ago
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    not just compute y

  67. anonymous
    • one year ago
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    wait isnt the inverse y = \[\sqrt{x+1} \] and not y= \[\sqrt{x-1}\]

  68. FibonacciChick666
    • one year ago
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    f(x)=x^2+1 y=x^2+1 switch x=y^2+1 x-1=y^2 sqrt(x-1)=y

  69. anonymous
    • one year ago
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    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
    • one year ago
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    to be onto, you must be able to hit every element of the range

  71. FibonacciChick666
    • one year ago
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    here our function is Z->[1,infty)

  72. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    This slideshow may help http://www.csee.umbc.edu/~stephens/203/PDF/7-3.pdf

  75. anonymous
    • one year ago
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    every function is onto its range, trivially :)

  76. anonymous
    • one year ago
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    not every function is onto its 'codomain'

  77. FibonacciChick666
    • one year ago
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    function implies surjective

  78. FibonacciChick666
    • one year ago
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    but you still have to learn to prove it

  79. FibonacciChick666
    • one year ago
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    it's part of proving something is a function.

  80. FibonacciChick666
    • one year ago
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    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
    • one year ago
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    ok. slide 7.3.6 is what i need.

  82. anonymous
    • one year ago
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    im looking at the slides so i might be gone a few minutes

  83. FibonacciChick666
    • one year ago
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    is cool

  84. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    you call the 'range' what i call the 'codomain'

  88. FibonacciChick666
    • one year ago
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    they are the same thing, but we didn't have one defined before

  89. FibonacciChick666
    • one year ago
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    ok, maybe not logically the exact same thing :P

  90. anonymous
    • one year ago
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    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
    • one year ago
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    I call it image. But to be honest, it's semantics to me. I learned range first it stuck.

  92. anonymous
    • one year ago
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    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
    • one year ago
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    sorry for the bad formatting but im slow using the equation and inserting them

  94. FibonacciChick666
    • one year ago
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    ok, now, leave out the middle bit. Just state a y value

  95. FibonacciChick666
    • one year ago
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    the actually finding the inverse is scratch work

  96. anonymous
    • one year ago
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    i guess 2, 3 or 4 any really

  97. anonymous
    • one year ago
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    oh well dang.

  98. FibonacciChick666
    • one year ago
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    oh, no I mean just claim y=inverse essentially

  99. FibonacciChick666
    • one year ago
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    you don't show the work there. You do have to show y is in your codomain/range though

  100. anonymous
    • one year ago
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    so just say if y is in Y the there is an x in Z (the integers) such that f(n) = y

  101. FibonacciChick666
    • one year ago
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    that's what you are proving.

  102. anonymous
    • one year ago
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    i thought i just proved it.

  103. FibonacciChick666
    • one year ago
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    "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
    • one year ago
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    just a restatement needed.

  105. anonymous
    • one year ago
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    ok. just for got 2 of the needed clarification statements. i see why i need them.

  106. FibonacciChick666
    • one year ago
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    Alright, so no promises on this being perfect, but I hope I helped

  107. anonymous
    • one year ago
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    i was excited and typing fast because the scratch work was where i was stuck

  108. anonymous
    • one year ago
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    you did

  109. FibonacciChick666
    • one year ago
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    I'm gonna try and nap an hour or two before my test

  110. anonymous
    • one year ago
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    you and jayzzd i really appreciate it. i like those slides.

  111. FibonacciChick666
    • one year ago
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    so, good morning/night and np

  112. anonymous
    • one year ago
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    test!! dang i would be asleep also

  113. FibonacciChick666
    • one year ago
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    praxis II math :( :( hence the not sleeping now

  114. anonymous
    • one year ago
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    i see well good luck.

  115. anonymous
    • one year ago
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    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
    • one year ago
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    example: f = { (1, 4 ) , ( 2 , 2) ( 3 , 7 ) } | f | = | { 1,2,3} |

  117. FibonacciChick666
    • one year ago
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    thanks, you too night all

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