# Let f : R → R be a function, such that $|f(x)−f(y)|≥5|x−y| \:\forall \:x, y\in \mathbb{R}$. Show that $f$ is injective. [closed]

Intro to Math Proofs course

Know basic concepts of Injection functions (one-to-one)

## closed as off-topic by Henrik, user10354138, Saad, A. Pongrácz, Kabo MurphyNov 22 '18 at 7:50

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• Welcome to MSE! What have you tried? – MisterRiemann Nov 21 '18 at 17:46
• Tried doing a contrapostive example (i.e for every x1, x2 in A x1 does not equal x2 implies that f(x1) does not equal f(x2) but i am not sure how to use this in this case with 2 variables – smith Nov 21 '18 at 17:50
• How about using the definition directly: Assume that $f(x)=f(y)$ and show that this implies $x=y$. – MisterRiemann Nov 21 '18 at 17:52
• That should work but I am unsure how to format or start the answer with 2 variables or how to use the inequality to show x=y. – smith Nov 21 '18 at 17:59
• Literally just write it, lol. If $f(x)=f(y)$ then literally write that in your inequality and see what you get. – user608030 Nov 21 '18 at 18:01

## 2 Answers

To prove that a funtion is injective you've got to show that $$f(x)=f(y) \Rightarrow x=y$$. Now try to prove this by contradiction, meaning if $$f(x)-f(y)=0$$ and $$x\neq y$$ that is $$|f(x)-f(y)|=0 \geq 5|x-y|$$, now, since $$x\neq y \Rightarrow |x-y|>0$$, which implies that $$0$$ is bigger than five times some positive integer, which is absurd. Now, it follows that is $$f(x)=f(y)$$ then $$x=y$$

• Very clear and helpful. Thank you – smith Nov 21 '18 at 18:19

Shortcut approach:
Given:$$|f(x)−f(y)|≥5|x−y| \:\forall \:x, y\in \mathbb{R}$$

By definition of injective functions: $$f(x)=f(y) \hspace{0.5cm}\text{implies} \hspace{0.5cm}x=y$$

So, if you put $$f(x)=f(y)$$, your LHS becomes $$0$$.
So needs to be RHS, which is only possible if $$x=y$$.
Proved!