1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proof Help: Proving one to one and onto

  1. Nov 17, 2005 #1
    Hello, I was having trouble proving that two functions are one to one and onto.

    For one:

    S = {x is an element of all reals: x = (n^2 + sqrt(2))/n, n is an element of naturals numbers}

    Define f: Natural Numbers -> S by f(n) = (n^2 + sqrt(2))/n

    We have to show that f is onto.

    So I believe what we have to do is show that there exists an n such that f(n) = t where t is an element of S for any t?

    So how would we go about that, do we solve the equation for t?

    t = (n^2 +sqrt(2))/n

    nt = n^2 +sqrt(2)

    n(t-n) = sqrt(2)

    ... but of course the small sequence we just went through proves absolutely nothing?

    I'm having a similar problem with proving that the function:

    For f: Natural Numbers -> set of all Integers

    f(r) = (1 + ((-1)^r)*(2r-1))/4

    is bijective

    For one to one I tried the contrapositive, set f(a) = f(b) for some a, b.

    So I set the equations equal to each other and simplified to:

    ((-1)^a)*(2a-1) = ((-1)^b)*(2b-1)

    but I can't see how to simplify this down to show that a=b...

    I'm having a similar problem with onto, using the same idea as we used for the first problem:

    Let q be an integer.

    q = f(n)

    4q - 1 = ((-1)^n)(2n-1)

    What can I show from here?

    Any help is appreciated, thanks.
  2. jcsd
  3. Nov 17, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Go back to the definition of S. What, precisely, does this notation mean?

    S := \left\{ x \in \mathbb{R} \, \left| \, \exists n \in \mathbb{N} : x = \frac{n^2 + \sqrt{2}}{n} \right\}

    In particular, if you know [itex]t \in S[/itex], what does this definition tell you?
  4. Nov 17, 2005 #3
    Thanks for the reply.

    You may have to help me out here a bit, as far as I know the notation means that t is an element of all real numbers such that it is equal to the equation for some natural number n. Thus every natural number has some t value, so we have to find the inverse of the function to solve for n, so then we can show that for any t we can obtain some n. I think this is what I’m supposed to do? But I’m not entirely sure if that's correct/how to do that, I tried solving for t and it didn't work, so I'm not entirely sure what your question is implying, unless I interpreted the question incorrectly?
  5. Nov 18, 2005 #4
    Regarding the first part you are probably over analyzing the question. S is the set of all reals FOR WHICH there exists an n in N SUCH THAT
    x=(n^2+sqrt(2))/n. Therefore S is the set
    S=(1+sqrt(2), (4+sqrt2)/2, (9+sqrt(2))/3, ....). To show that f is surjective you must show that for any s in S, there exists an n in N, such that s=f(n). By the definition of f this turns out to be rather trivial.
    Pick an arbitrary s in S, then we know there exists an n in N such that s=(n^2+sqrt(2))/n, but this is precisely the definition of f(n). Therefore the mapping is surjective.
  6. Nov 18, 2005 #5
    for the 1st part [tex]f(\sqrt{ny-\sqrt{2}}) = x[/tex] & [tex]f(-\sqrt{ny-\sqrt{2}}) = x[/tex], so x is the image of some element of that set S, showing that f is onto.

    the 2nd one is similar. & just set f(x)=f(y) & do some algebra to get x=y.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook