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Hey guys, I've a small algebraic problem about quantifiers:
I was looking through my hmwk today, and it says i must express: "There is no smallest positive real number" using quantifiers:
so far i got:
(1) ∀x ∃y (x>y)
For any x, there exists a y which is smaller.
(2) ∃x ∀y (x<y)
There exists an x such that for all y, x is smaller than y.
I think the correct quantifier statement is (1). But my friend said i'm wrong...so now I'm slightly confused. Isnt (2) saying that there exists an x thats smaller than ANY y? which means there IS a smallest positive real number right?
Hope you guys can enlighten me =/ I'm confuseddd
Thanks guys, I think I get it now 
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Belgium9947 Posts
You're right.
And the second sentence says exactly what you think it does.
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Yeah your friend is no match for TL.net wisdom :D
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∀x ∃y (y<x ^ y > 0)
x,y ∈ R
i guess
edit:
(2) ∃x ∀y (x<y)
There exists an x such that for all y, x is smaller than y.
this implies x can be negative too.
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Belgium9947 Posts
On September 30 2009 01:22 EsX_Raptor wrote:∀x ∃y (y 0)
x,y ∈ R
i guess
edit:
this implies x can be negative too.
...
lol
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the 2nd one almost works...
here's what it should be
!∃x>0 ∈ R ∀y>0 ∈ R (x<y)
translation:
There does not exist a real number x greater than zero such that for all real numbers y greater than zero x is less than y.
Essentally, there's no number that is less than every other number in the set of positive real numbers.
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Belgium9947 Posts
Oh wow I didn't notice that it had to be positive LOL sorry
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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?
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I would use R+ to make the notation simpler.
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Although some of the answers here are equivalent, I also think Caldo's answer using the "not exist" quintifier would be the direct translation, not just an equivalent statement .
Edit: Just interested, why do you consider it an algebraic problem  ? I always think of quantifiers as a part of analasys, maybe cause that was the first time I saw them . Also, they're most commonly used in analasys imo.
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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."
On September 30 2009 04:00 Papvin wrote:Just interested, why do you consider it an algebraic problem ?
I was wondering this too... I thought algebra was like factoring and equation manipulation and whatnot, not quantifiers and sets so much.
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How do you integrate [ln(x^2 + 1) dx] using IBP?
By IBP, I could get integral (lnx dx), to xlnx - x + C
But, with ln (something something), not just lnx. I can't seem to do it.
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Germany2896 Posts
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Algebra is the reason i quitted maths. No goals and no links with real world made me hate it. Also it seems that all the algebra teachers are either retarded or weirdos.
Sry if i have offended anyone. I wish you good luck and i hope you enjoy it.
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dude whats up with the wierd symbols? -_-
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On September 30 2009 06:01 Boblion wrote:
Algebra is the reason i quitted maths. No goals and no links with real world made me hate it.
This may well be true for you. Still, for better or for worse that is not the case for many people. Any technical / quantitative field will use equations and algebra.
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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.
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^ it's generally understood that R is the set of all real numbers, so by stating x∈R we're saying that x is in the set of all real numbers,which implies that x is a real number.
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and this is why I hate maths
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+ 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?
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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
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Ahh can't wait to get to modern algebra..
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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.
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EPT
