Hello, I was having trouble proving that two functions are one to one and onto.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Proof Help: Proving one to one and onto

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