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I am working on this question first I want to understand the question itself, what was the question asking me?

For me, I think $\mathbb{R}$ to $\mathbb{R}$ are real numbers and if $\mathbb{R}$ to $\mathbb{R}$ is defined as $f(x)= \cos x$, then I must prove that $\cos x$ is an onto function if it maps from real numbers $\mathbb{R}$ to $\mathbb{R}$.

Can someone elaborate on the question? I will really appreciate that.

N. F. Taussig
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Surdz
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    The question is asking if every real number is in the range of the cosine function. – N. F. Taussig May 23 '16 at 01:08
  • can you simply elaborate on what is a range? – Surdz May 23 '16 at 01:33
  • The range of a function $y = f(x)$ is the set of all $y$-values it assumes. The function $f(x) = \cos x$ has range $[-1, 1] = {y \in \mathbb{R} \mid -1 \leq y \leq 1$ since the cosine function assumes all values from $-1$ to $1$ inclusive and only those values. – N. F. Taussig May 23 '16 at 01:36
  • But what does it mean when real numbers are beyond this range. since real numbers are beyond this range. – Surdz May 23 '16 at 01:44
  • If there are real numbers that are not in the range of the function, then the function is not surjective. If the cosine function were surjective, then every real number would be in its range. Since this is not the case, you are supposed to conclude that $f(x) = \cos x$ is not surjective, which is why Olivier Oloa provided the hint that he did. – N. F. Taussig May 23 '16 at 01:49
  • Let us continue this discussion in chat. – Surdz May 23 '16 at 02:38

4 Answers4

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Hint. One may recall that, for all $x \in \mathbb{R}$, $$ -1\leq \cos (x) \leq 1. $$ Then do you think it is possible to find a real number $x$ such that for example $$\cos(x)=2\,?$$

Olivier Oloa
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Let $f$ be a function from set $A$ to set $B$, denoted $f: A \to B$. The set $A$ is called the domain of $f$. The set $B$ is called the codomain of $f$. The set $f(A) = \{f(a) \mid a \in A\}$ is called the range of $f$. A function is said to be surjective when the codomain is equal to the range, that is, if $f(A) = B$.

Example. Let $A = \{a, b, c, d\}$. Let $B = \{1, 2, 3, 4\}$. The function $f: A \to B$ defined by $f(a) = 1$, $f(b) = 2$, $f(c) = 3$, $f(d) = 4$ that sends each letter to its position in the alphabet is surjective since $f(A) = \{f(a), f(b), f(c), f(d)\} = \{1, 2, 3, 4\} = B$.

Example. Let $A = \{a, b, c, d\}$. Let $B = \{1, 2, 3, 4\}$. The function $g: A \to B$ defined by $g(a) = 1$, $g(b) = 2$, $g(c) = 3$, $g(4) = 1$ is not surjective since $g(A) = \{1, 2, 3\} \subsetneq B$ because the element $4 \in B$ is not in the range $g(A)$.

A function $f: \mathbb{R} \to \mathbb{R}$ has domain and codomain equal to the set of all real numbers. It is surjective if the range of the function is also the set of all real numbers.

An example of a surjective function $f: \mathbb{R} \to \mathbb{R}$ is the function $f(x) = x^3$ since its range is $\mathbb{R}$.

cube_function_graph

An example of a function $f: \mathbb{R} \to \mathbb{R}$ that is not surjective is $f(x) = x^2$ since its range $[0, \infty) = \{x \in \mathbb{R} \mid x \geq 0\}$ does not include any negative real numbers.

square_function_graph

The question is asking you whether the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \cos x$ is surjective, that is, it is asking you whether the range of $f(x) = \cos x$ is the set of all real numbers.

cosine_function_graph

N. F. Taussig
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if a F map from a to b (in this case the reals to the reals) and is surjective or "onto" means that for every element x in b: x is hit by an element y in A. or inother words for all x there exist a y such that F (y)=x. some good examples and illustrations are here surjective functions

shai horowitz
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To visualize this metaphorically, imagine that you take every element of your domain, apply the function, and place a marker or sticker or poker chip... whatever... onto or on the surface of the mapped image in the codomain. Have you completely covered your codomain, $\mathbb{R}$, or are there any numbers the mapping fails to reach?

If you've completely covered the codomain, it is surjective.

zahbaz
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