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Prove that the function $f: R\to R$ given $f(x)=x\cdot |x|$ is a bijection

Proof 1: f is injective

Suppose $$f(a) = f(b) \implies a\cdot |a| = b\cdot |b|$$ $$\implies (a\cdot |a|)^2 = (b\cdot |b|)^2$$ $$\implies a^4 =b^4$$ $$a=b$$ Thus, f is injective

Proof 2: f is surjective

Pick $n\in R$, we want to show $n=f(x)$ for some $x\in R$

$$\implies n= a\cdot |a|$$ for some $a\in R$

I dont know how to complete the surjective proof

TheGamer
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    $a^4=b^4$ does not imply that $a=b$. – lulu Jul 21 '17 at 19:25
  • @lulu but since they have same exponent, i thought a=b – TheGamer Jul 21 '17 at 19:27
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    No. $(-1)^4=1^4$, for example. You need to modify your argument to take signs into account (i.e. first assume $a,b>0$ then $a>0,b<0$ and so on.) – lulu Jul 21 '17 at 19:27
  • "but since they have same exponent, i thought a=b" Why? Does $x^4 = k$ have only one solution? What if $a$ is one of the solutions and b is the other. in general squaring both sides of an equation ADD extra false solutions. – fleablood Jul 21 '17 at 19:38
  • @fleablood so $ x\neq y$, thus the function is not bijection – TheGamer Jul 21 '17 at 19:43
  • No. It is a bijection. But your proof is wrong. – fleablood Jul 21 '17 at 19:56
  • Notice $a|a| \ge 0 \iff a \ge 0$ and $b|b| \ge 0 \iff b \ge 0$ so $a|a| = b|b|$ only if $a$ and $b$ have the same sign. So if you can prove $|a| = |b|$ then in this case, since $a$ and $b$ have the same sign, that would mean $a = b$. – fleablood Jul 21 '17 at 19:59
  • You do have to know that a. For every $b \ge 0$ there is a unique $c = \sqrt{b} \ge 0$ so that $c^2 = b$. That's probably being given to you as a given or maybe that has been proven. But you will need that. Equivalently you may have $a^{2n} = b^{2n} \implies |a| = |b|$. You can use that. But you can NOT use $a^{2n} = b^{2n} \implies a = b$ because that is FALSE. – fleablood Jul 21 '17 at 20:03

4 Answers4

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Observe that for each real $x $,

$$f (-x)=-f (x) $$ $f $ is an odd function $(f (0)=0) $.

For $x>0 , \; f (x)=x^2 \;\;,f'(x)=2x. $

$f $ is continuous at $(0,+\infty) $ and strictly increasing. thus $f $ is a bijection from $(0,+\infty) $ to $(\lim_{0^+}x^2=0,\lim_{+\infty}x^2=+\infty) $ hence $f $ is a bijection from $\Bbb R $ to $\Bbb R $.

hamam_Abdallah
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If $y\ge 0$ then $y=f(\sqrt y)$. If $y<0$ then $y=f(-\sqrt{-y})$.

Hagen von Eitzen
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The given function is $$f(x)= \begin{cases} x^2 & \text{ if } x \geq 0\\ -x^2 & \text{ if } x < 0\\ \end{cases}$$ Let $a \in \mathbb{R}$.

If $a\geq 0$, then we can choose $x=\sqrt{a}$ to get $f(x)=a$. If $a< 0$, then we can choose $x=-\sqrt{-a}$ to get $f(x)=a$.

So we can map to any real number. Hence surjective.

Note: Also your proof for injectivity needs fixing. This representation of the function can help you with that as well.

Anurag A
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  • $x\cdot |x| =x^2$? – TheGamer Jul 21 '17 at 19:35
  • @TheGamer the reason for that is $|x|=x$, when $x \geq 0$ and $|x|=-x$, when $x<0$. – Anurag A Jul 21 '17 at 19:36
  • |x| = x if x >= 0 so x|x| = x^2 if x >= 0. |x| = -x is x < 0 so x|x| = -x^2 if x < 0$.. – fleablood Jul 21 '17 at 19:40
  • @AnuragA is this right for injective: Case1: x,y>0 $$\implies x^2=y^2$$ $$x=y$$ Case1: x,y<0 $$\implies -x^2=-y^2$$ $$x=y$$ – TheGamer Jul 21 '17 at 20:47
  • @TheGamer actually you should start with $f(x)=f(y)$, this means both $f-$values have the same sign, so based on the piecewise function I have defined you know that either $f(x)=f(y) \implies x^2=y^2$ (i.e both $x,y \geq 0$) or it $\implies -x^2=-y^2$ (both $x,y <0$) In either case you have $x^2=y^2 \implies x =\pm y$ but based on the domain of the function we know that either both inputs $x,y \geq 0$ or both $x,y <0$. Thus the only possibility is $x=y$. – Anurag A Jul 21 '17 at 22:44
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$a^4 = b^4 \not \implies a = b$ as $x^4 = k>0$ will have two real solutions $\sqrt[4]{k}$ and $-\sqrt[4]{k}$ so it will be possible that $a = -b \ne b$ yet $a^4 = (-b)^4 = b^4$.

So that's no good.

Indeed the ONLY reason one should square both sides to solve an equation is to allow for opposite signs of values to become equivalent and to be ignored. Squaring both sides ADDS extra, not necessarily correct "solutions".

$x = 5$ has one solution.

$x^2 = 5^2$ has two solutions.

$x = 5$ OR $x = -5$. $x = -5$ is an added extraneous solution. Watch out for those.

.....

Anyway....

Surjection is the easy part.

Let $w \in \mathbb R$

Let $sign(w) = -1$ if $w < 0$ and $sign(x) = 1$ if $w \ge 0$.

Let $x = sign(w)*\sqrt{|w|}$.

The $|x| = \sqrt{|w|}$ and $x*|x| = (sign(w))*\sqrt{|w|}*\sqrt{|w|} = sign(w)*\sqrt{|w|^2} = sign(w)*|w| = w$.

Now go back and fix your proof that it is injective.

fleablood
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  • fleablood whats sign? – TheGamer Jul 21 '17 at 20:01
  • I just defined it. sign(w) = -1 if w < 0. And sign(w)= 1 if w>=0. It's a function I just invented. Basically if $w$ is a negative number than $sign(w) = -1$. and if $w$ is not negative then $sign(w) = 1$. I figured that would be useful. So I defined it and used it. Point being $w = sign(w) *|w|$. – fleablood Jul 21 '17 at 20:07
  • In other words. If $w \ge 0$ let $x = \sqrt{w}$. If $w < 0$ let $x = -\sqrt{|w|}=\sqrt{-w}$. then if $w \ge 0$, $x|x| = \sqrt{w}\sqrt{w} = w$. If $w < 0$ then $x|x| = -\sqrt{|w|}|\sqrt{|w|} = -|w| = w$. – fleablood Jul 21 '17 at 20:09