How to prove one-one function

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How to prove one-one function

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Vulch

#1 h Apr 11, 2025, 8:03 PM

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Hello everyone,
I am learning functional equations.
To prove the below problem one -one function,I have taken two non-negative real numbers $(1,2)$ from the domain $\Bbb R_{*},$ and put those numbers into the given function f(x)=1/x.It gives us 1=1/2.But it's not true.So ,it can't be one-one function.But in the answer,it is one-one function.Would anyone enlighten me where is my fault? Thank you!

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This post has been edited 3 times. Last edited by Vulch, Apr 11, 2025, 8:07 PM

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#2 h Apr 11, 2025, 8:06 PM

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allright

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#3 h Apr 11, 2025, 8:35 PM

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(1,2) is not a point on the function

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#4 h Apr 11, 2025, 9:02 PM

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Do you know what one-one and onto functions are?

By definition one-one functions, also known as injective functions, are functions such that for any $2$ distinct inputs, they cannot give the same output.

For example,
Say $f: A \longrightarrow B$ is a function. $a,b \in A$ . If $f$ is one-one then for any choice of $a$ and $b$ such that $a \neq b$ , $f(a)$ cannot be equal to $f(b)$ , that is, $f(a) \neq f(b)$ .

So for example in your example, $f(1) = \frac{1}{1}$ and $f(2) = \frac{1}{2}$ . As $f(1) \neq f(2)$ , $(a,b) = (1,2)$ is not a correct example to contradict injectivity of the function.

In this prove, we will FTSOC assume that the function is not one-one. So there must exist a pair $(a,b) \in {\mathbb{R}^{+}}^2$ such that $a \neq b$ and $f(a) = f(b)$ . But that means that $\frac{1}{a} = \frac{1}{b}$ . And since both $a,b$ are not $0$ , $a=b$ . This contradicts our assumption and hence $f$ must be injective.

Now can you show its surjectivity?

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Vulch

#5 h Apr 12, 2025, 2:07 AM

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Would you show its subjectivity?

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#6 h Apr 12, 2025, 2:56 AM

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Vulch wrote:

Would you show its subjectivity?

Let $x\in\mathbb R*$ be any nonzero real number, for surjectivity we need to show that there is a $y\in\mathbb R*$ with $f(y)=x$ . We can choose $y=\frac1x$ , since clearly $f\left(\frac1x\right)=x$ and $\frac1x\in\mathbb R*$ .

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Vulch

#7 h Apr 16, 2025, 2:20 PM

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Solve the following problem:

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#8 h Apr 18, 2025, 1:23 PM

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Vulch wrote:

Solve the following problem:

What have you done so far?

Of course it is not one-one as $f(0.5) = f(1) = 1$ .
Of course it is not onto as there is no $k$ such that $f(k) = 0.5$ .

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