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Intro to Math Proofs course

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

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

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  • $\begingroup$ Welcome to MSE! What have you tried? $\endgroup$ – MisterRiemann Nov 21 '18 at 17:46
  • $\begingroup$ 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 $\endgroup$ – smith Nov 21 '18 at 17:50
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    $\begingroup$ How about using the definition directly: Assume that $f(x)=f(y)$ and show that this implies $x=y$. $\endgroup$ – MisterRiemann Nov 21 '18 at 17:52
  • $\begingroup$ 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. $\endgroup$ – smith Nov 21 '18 at 17:59
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    $\begingroup$ Literally just write it, lol. If $f(x)=f(y)$ then literally write that in your inequality and see what you get. $\endgroup$ – user608030 Nov 21 '18 at 18:01
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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 $x=y$

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  • $\begingroup$ Very clear and helpful. Thank you $\endgroup$ – smith Nov 21 '18 at 18:19
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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!

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