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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


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    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.
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