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I think that the statement is true in general considering +1 or -1 for y. How can I prove it in proper notation.

Similarly I need to prove $ \exists y \in \mathbb R, \forall x \in \mathbb R st, x^2 +2x - 5 \leq y$

I did it as follows, is it wrong?

$$x^2 +2x - 5= (x+1)^2 -6 \leq -6 \forall x \in \mathbb R$$ $$ \therefore \exists y \in \mathbb R, \forall x \in \mathbb R st,x^2 +2x - 5 \leq y$$ is true.

S.Dan
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  • As for $|x|$, use the definition of $|.|$ and you are done. The second claim is just false, as is your claim towards a proof. – Stefan Mesken Aug 11 '14 at 16:13
  • write out your definition of $| \cdot |$ as a function first. It's a piecewise definition, defined differently for when $x \ge 0$ and $x<0$, so you consider these two cases. In the second part of the question I think you're confused, the statement you've written is false as it stands. – JC574 Aug 11 '14 at 16:13
  • The equal sign is not supposed to be there, Eddited the question. – S.Dan Aug 11 '14 at 16:23
  • Second claim is still wrong since the parabola opens up. – G. Bach Aug 11 '14 at 16:27
  • If the sratement is " $\geq $ ", would it be correct? – S.Dan Aug 11 '14 at 16:45
  • @S.Dan yes, it would. You could also switch the order of the quantifier statements. – Ben Grossmann Aug 11 '14 at 16:52

2 Answers2

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$$|x|=\begin{cases}+x,\;x\ge0\\-x,\;x\le0\end{cases}$$ So,$$y=\frac{|x|}x=\begin{cases}+1,\;x>0\\-1,\;x>0\end{cases}$$ And for $x=0$, $|x|=xy\forall y\in\mathbb R$

$y$ can also be written as $$y=\mathit{sgn}(x),\;\forall x\ne0\wedge y\in \mathbb{R},\;x=0$$

Actually, $$\exists y \in \mathbb R, \forall x \in \mathbb R \text{ such that } x^2 +2x - 5 \geq y$$ Since, $$x^2 +2x - 5= (x+1)^2 -6 \ge -6 \qquad \forall x \in \mathbb R $$ Since, $$(x+1)^2\ge 0 \qquad \forall x \in \mathbb R$$

RE60K
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  • It is slightly odd that nobody (in any case, neither the OP nor the upvoters) saw fit to note that $$\frac{|x|}x$$ is undefined when $x=0$, hence the post does not answer the question. – Did Aug 11 '14 at 21:59
  • @Did does the edit makes it fit? – RE60K Aug 12 '14 at 00:25
  • Not quite since now there is no definition of $y$ when $x=0$. – Did Aug 12 '14 at 00:30
  • Likewise, the whole post after "Actually" does not seem to follow the rules of deduction and quantification, perhaps because of the slightly odd "such that". – Did Aug 12 '14 at 08:19
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$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & } \newcommand{\endcalc}{\end{align}} $Here is another way to write down the proof for your first question. I'm using a slightly different notation that I personally think is better for proofs like this.

(Note that this is not "proper notation", since I don't think that exists. There are only different conventions, preferences, and opinions. I happen to like those from EWD1300.)

For all $\;x \in \mathbb R\;$,

$$\calc \langle \exists y : y \in \mathbb R : |x| = x \times y \rangle \calcop{\equiv}{divide by $\;x\;$, with special case $\;x = 0\;$ using $\;|0| = 0 \times y\;$} \langle \exists y : y \in \mathbb R : x = 0 \;\lor\; \frac{|x|}x = y \rangle \calcop{\equiv}{logic: pull part not using $\;y\;$ out of $\;\exists y\;$} x = 0 \;\lor\; \langle \exists y : y \in \mathbb R : \frac{|x|}x = y \rangle \calcop{\equiv}{logic: one-point rule for $\;y\;$} x = 0 \;\lor\; \frac{|x|}x \in \mathbb R \calcop{\equiv}{arithmetic: right-hand side is always true (for $\;x \not= 0\;$); simplify} \text{true} \endcalc$$

MarnixKlooster ReinstateMonica
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