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help with algebra :( - Page 2

Blogs > Mr.Maestro
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EsX_Raptor
Profile Blog Joined February 2008
United States2802 Posts
Last Edited: 2009-09-29 23:52:29
September 29 2009 23:50 GMT
#21
On September 30 2009 05:10 caldo149 wrote:
Show nested quote +
On September 30 2009 03:11 EsX_Raptor wrote:
caldo is right, that's the answer.

edit: you made me think, you clearly state there is no smallest positive real number while i (somewhat) state there is always a smaller number (which also means there is no smallest one). I guess they're somewhat equivalent?

My fixed version should be:

∀x∈R ∃(y>0)∈R (y<x)

Any thoughts?

I think that your expression would be equally valid if you stated x>0, otherwise i can think of examples that make it false easily. With that quick fix though, our solutions both imply the same things and solve the given problem. I was just a bit more literal with my "translation."

Oh I see, didn't notice that! Thank you for your response n_n this had me confused for a while haha

edit: for those who say math sucks, you haven't really gotten well into it! It can get pretty fascinating after a while
Dave[9]
Profile Blog Joined October 2003
United States2365 Posts
September 30 2009 01:38 GMT
#22
Ahh can't wait to get to modern algebra..
http://www.teamliquid.net/forum/viewmessage.php?topic_id=104154&currentpage=316#6317
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
September 30 2009 01:47 GMT
#23
you are right.
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
IMlemon
Profile Blog Joined May 2008
Lithuania296 Posts
September 30 2009 07:33 GMT
#24
On September 30 2009 08:31 Papvin wrote:
+ Show Spoiler +
On September 30 2009 06:25 IMlemon wrote:
Show nested quote +
On September 30 2009 03:11 EsX_Raptor wrote:
caldo is right, that's the answer.

edit: you made me think, you clearly state there is no smallest positive real number while i (somewhat) state there is always a smaller number (which also means there is no smallest one). I guess they're somewhat equivalent?

My fixed version should be:

∀x∈R ∃(y>0)∈R (y<x)

Any thoughts?


Problem with this, is that it's not a valid formula in mathematical logic. Can you use predicates? If so, something like this would do.

P(x) - number is real
R(x) - number is positive
Q(x,y) - y is smaller than x

F = ∀x∃y (P(x) /\ P(y) /\ R(x) /\ R(y) /\ Q(x,y))

^True if x is positive and real, false otherwise. y must be kept in check too.

If you can't use predicates, im kinda out of ideas how to express it precisely. To state that x ∈ R you'd have to write out all of the real numbers' properties.

Instead of your F, setting R+ to the set of real (strictly) positive numbers, wouldn't it suffice to write
F = ∀x∈R+∃y∈R+:y<x?
Or are you speaking of stricly formal mathematical language, where nothing is left to the intuition?


I assume OP wanted to get a valid formula. Thingies caldo wrote above aren't legit. If you compare that to equations, it would be the same thing as writing, say " = x(x > 9) , sqrt (3 < y)". While it's obvious what you mean, it's wrong in mathematical (gay) sense.

Math gets stupidly abstract and boring really fast.
My future's so bright, I gotta wear shades.
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