• Let $f:\mathbb{N}\rightarrow\mathbb{Z}$ be a function which $f(n)=\dfrac {(-1)^n (2n-1)+1} {4}$. Show that $f$ is surjection.

Proof. Let $m\in\mathbb{Z}$. We need to find $n\in\mathbb{N}$ such that $f(n)=m$.Let $m=\dfrac {(-1)^n (2n-1)+1} {4}$ for $n\in\mathbb{N}.$ If $n$ is even then we get

$$m=\dfrac {(-1)^n (2n-1)+1} {4},$$



If $n$ is odd, then we get



Therefore, can we say $f$ is surjective?


You can leave it more explicit

Let $n \in \mathbb{N}$

If $n$ is even, $n = 2k$ for some $k \in \mathbb{N}$

Then $f(2k) = f(n) =\frac{(-1)^{n}(2n-1)+1}{4} = \frac{(2n-1)+1}{4} = \frac{2n}{4} = \frac{4k}{4} = k $

So, for each $m \in \mathbb{N}$, if we take $n = 2m$, we have that $f(n) = f(2m) = m$

Now if $n$ is odd, we have $n = 2k -1$ for some $k \in \mathbb{N}$

Then $f(2k-1) = f(n) = \frac{(-1)^{n}(2n-1)+1}{4} = \frac{(-2n +1) +1}{4} = \frac{(-2n + 2)}{4} = \frac{-2(2k-1) +2}{4} = \frac{-4k + 4}{4} = -k + 1$

So if $m \in \mathbb{Z}, m ≤ 0$ then $m - 1≤-1$ so $-m +1≥ 1$ Then $-m + 1 \in \mathbb{N}$ And if we take $n = 2(-m+1) -1$ then we have $f(n) = f(2(-m+1)-1) = -(-m+1) + 1 = m$

Then $f$ is surjrctive.


Your proof is a little backward. You need to find $n \in \mathbb{N}$ such that $f(n)=m$. So you can't say "let $m= f(n)$", because you don't know that there is such $n$ a priori. As such, it doesn't make sense to say "if $n$ is even" or "if $n$ is odd", since you have not demonstrated that $n$ exists. What you have written down is like the scratch work to figure out how the proof should go, but it does not constitute a proof.

Notice that: \begin{align*} f(1)&=0\\ f(2) &= 1 \\ f(3) &= -1\\ f(4) &=2\\ f(5)&=-2 \\ &\vdots \end{align*}

Now consider separately the cases $m=0$, $m>0$, and $m<0$. In each case, come up with a formula for $n$ in terms of $m$ that will satisfy $f(n)=m$.


a little algebra shows: $$ f(n+2) - f(n) = (-1)^n $$

since $f(0) = 0$, then by a telescoping sum: $$ f(0+2m) = \sum_{k=1}^m (-1)^0 = m $$

similarly, $f(1) = 0$, and so $$ f(1+2m) = \sum_{k=1}^m (-1)^1 = -m $$


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