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kevoBest ResponseYou've already chosen the best response.0
Prove that \[\exists z \in \mathbb{R} \forall x \in \mathbb{R}^+[\exists y \in \mathbb{R}(yx+y/x) <> x \neq z)\]
 one year ago

kevoBest ResponseYou've already chosen the best response.0
Lol, sorry for the wait, it was a pain to type up.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.0
What does <−−> mean?
 one year ago

kevoBest ResponseYou've already chosen the best response.0
its the symbol for if and only if.
 one year ago

eliassaabBest ResponseYou've already chosen the best response.0
It is still not clear to me what the last statement mean. \[ \exists z \in \mathbb{R} \forall x \in \mathbb{R}^+[\exists y \in \mathbb{R} \] such that what?
 one year ago

Wheaton71Best ResponseYou've already chosen the best response.0
Yeah I'd also like to know.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
just fyi, if you want to increase the space between characters in the equation editor, simply type "\;" for a small space, "\:" for a slightly bigger one, "\quad" for a big one, and "\qquad" for a giant one. This helps increase readability.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
I think the best way to approach this would be to split it into two parts. First we want to show implication tot he right, and second we want to show implication to the right. Also note that \[p \Rightarrow q \quad \Longleftrightarrow \neg \;p \;\;\text{V}\;\;q\]
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
So to show implication to the right, let's see if we can prove the simpler statement.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
I'm a little confused about the statement \((yx+y/x)\). What is it saying? In this form it's virtually meaningless.
 one year ago

kevoBest ResponseYou've already chosen the best response.0
That's a good question.. so there exists y in R(y  x = y/x) iff x does not equal z.
 one year ago

kevoBest ResponseYou've already chosen the best response.0
This is just one confusing statement that should not be legal to give to student. Just saying.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
There's supposed to be an equals sign there. That helps. Give me a second to think about this.
 one year ago

kevoBest ResponseYou've already chosen the best response.0
LOL i just realized that I mistyped that. Sorry!
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
Let's show implication to the right first. To show this, we need to choose a z such that for all y and x (x positive) \((yx =y/x)\) or \(x\neq z\). Just choose z to be negative. Since x is positive, we know that \(x \neq z\).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
Now we need to show implication to the left. To show this, we need to choose a z such that for all positive x, there exists a y such that that \(z=x\) or that \(yx\neq y/x\) Here, just choose \(y=0\). Since x is positive, \(0x\) is less than 0, and \(0/x=0\).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
Therefore, we are done. Sorry that took a while for me to write. Also, Instead of writing "we need to choose a z such that for all y and x (x positive)" in the first part, I should have written "we need to choose a z such that for all positive x there exists a y" It doesn't really matter in the end however.
 one year ago

kevoBest ResponseYou've already chosen the best response.0
KingGeorge. You are amazing. You are seriously my hero. I don't know how you are so good at this, but thank you.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
Practice, and a little bit of natural skill is how I'm good. I've also had some amazing teachers.
 one year ago

kevoBest ResponseYou've already chosen the best response.0
I know who I'm asking for help on proofs from now on :). I just don't know how I can reward you..
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.2
As long as you're trying to learn, I'll be good.
 one year ago

kevoBest ResponseYou've already chosen the best response.0
Well, if you're ever in Seattle, I'll buy you dinner.
 one year ago
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