Simple question 0<|x+3|<1/4
The notation of the book (its a paperback) I'm reading gives the answer of
-13/4<x<-11/4
And I was thinking "Where's the 'x does not equal -3' at?"
So I used Mathematica but it gave pretty much a spew of an answer
http://img3.imageshack.us/img3/1231/70441947.jpg
I want to know if I'm right and the book is wrong, because x cannot equal -3 in order for the argument to be greater than 0.
I think I'm looking for the correct notation of the answer more or less. Would it be the "-13/4<x<-11/4 INTERSECTION x does not equal -3"? Pickyness can win me points and slopiness deducts them, this is why I ask.
Just for the know. Can I ask more complicated questions in the future? The ones that I really would ned help with.
Since if and only if x = −3, then | x + 3 | = 0 then there are one of three four possibilities.
Deleting discussion about fate of thread now that the final edits are in.
Mathmatica is fun.
0 < |x + 3| < ¼
0 ≤ |x + 3| < ¼
- Your book's author intended to write 0 ≤ | x + 3 | < ¼ and either the author, editor, or the printer put in < where ≤ "less than or equal to" was intended.
- Your book's author intended to write x ≠ −3 in the solutions section, and either the author, editor, or the printer skipped over it or deleted it.
- Your author's a possibily malicious idiot, because 0 ≤ | z | is already obeyed for all complex z, so if they wrote 0 < | z | then they of course imply that z ≠ 0, and the editor, if there is one, failed to catch it.
- Your copy or eyes are old and the original has been distorted.
Deleting discussion about fate of thread now that the final edits are in.
Mathmatica is fun.
0 < |x + 3| < ¼
0 ≤ |x + 3| < ¼
Than you.
Simply for the "or.' I do think I recollect 'and' can refer to a union of subsets and 'or' can refer to an intersection. A simple or can maybe buy me some time in a final exam.
Simply for the "or.' I do think I recollect 'and' can refer to a union of subsets and 'or' can refer to an intersection. A simple or can maybe buy me some time in a final exam.
Actually is it's the opposite.
A < x is a upper half-line
x < B is a lower half-line
A < x < B is a finite segment but we can always represent it as the intersection of two half-lines.
A < x and x < B ⇔ A < x < B
In the same way, a union of finite segments can be represented with "or"
A < x < B or C < x < D
So for something in the form:
0 <= A < | x + B | < C we have:
− C − B < x < − A − B or A − B < x < C − B
But 0 <= A < | x + B| has the solution in the form of the union of two (disjoint) half-lines:
x < − A − B or A − B < x
A < x is a upper half-line
x < B is a lower half-line
A < x < B is a finite segment but we can always represent it as the intersection of two half-lines.
A < x and x < B ⇔ A < x < B
In the same way, a union of finite segments can be represented with "or"
A < x < B or C < x < D
So for something in the form:
0 <= A < | x + B | < C we have:
− C − B < x < − A − B or A − B < x < C − B
But 0 <= A < | x + B| has the solution in the form of the union of two (disjoint) half-lines:
x < − A − B or A − B < x
QUOTE (rpenner+Mar 5 2009, 07:28 AM)
Actually is it's the opposite.
A < x is a upper half-line
x < B is a lower half-line
A < x < B is a finite segment but we can always represent it as the intersection of two half-lines.
A < x and x < B ⇔ A < x < B
In the same way, a union of finite segments can be represented with "or"
A < x < B or C < x < D
So for something in the form:
0 <= A < | x + B | < C we have:
− C − B < x < − A − B or A − B < x < C − B
But 0 <= A < | x + B| has the solution in the form of the union of two (disjoint) half-lines:
x < − A − B or A − B < x
I don't know if this is a prank... (elpisis means I was drunk)
You contradict what you posted (mathematica) as an aswer.
A < x is a upper half-line
x < B is a lower half-line
A < x < B is a finite segment but we can always represent it as the intersection of two half-lines.
A < x and x < B ⇔ A < x < B
In the same way, a union of finite segments can be represented with "or"
A < x < B or C < x < D
So for something in the form:
0 <= A < | x + B | < C we have:
− C − B < x < − A − B or A − B < x < C − B
But 0 <= A < | x + B| has the solution in the form of the union of two (disjoint) half-lines:
x < − A − B or A − B < x
I don't know if this is a prank... (elpisis means I was drunk)
You contradict what you posted (mathematica) as an aswer.
I'm only contradicting your use of and/or versus intersection/union. You got the connection backwards.
QUOTE (rpenner+Mar 5 2009, 07:38 PM)
I'm only contradicting your use of and/or versus intersection/union. You got the connection backwards.
Your wrong.
BTW I never like linguistic use of mathematics, I can use a big upside down U instead.
Your wrong.
BTW I never like linguistic use of mathematics, I can use a big upside down U instead.
QUOTE (Beer w/Straw+Mar 5 2009, 07:43 PM)
Your wrong.
You're wrong
Sorry, I'll leave you two to your discussion
You're wrong
Sorry, I'll leave you two to your discussion
http://en.wikipedia.org/wiki/Intersection_(set_theory)
"x is an element of A ∩ B if and only if x is an element of A and x is an element of B."
http://en.wikipedia.org/wiki/Logical_conjunction
"The intersection used in set theory is defined in terms of a logical conjunction: x ∈ A ∩ B if and only if ( x ∈ A ) ∧ ( x ∈ B ). "
http://us.metamath.org/mpegif/elin.html
"Expansion of membership in an intersection of two classes. Theorem 12 of [Suppes] p. 25.
⊢ (A ∈ (B ∩ C) ↔ (A ∈ B ⋀ A ∈ C)) "
----
http://en.wikipedia.org/wiki/Union_(set_theory)
"Formally, x is an element of A ∪ B ∪ C if and only if x is in A or x is in B or x is in C."
http://en.wikipedia.org/wiki/Logical_disjunction
"The union used in set theory is defined in terms of a logical disjunction: x ∈ A ∪ B if and only if ( x ∈ A ) ∨ ( x ∈ B ). "
http://us.metamath.org/mpegif/elun.html
"Expansion of membership in class union. Theorem 12 of [Suppes] p. 25.
⊢ (A ∈ (B ∪ C) ↔ (A ∈ B ⋁ A ∈ C))"
Suppes, Patrick, Axiomatic Set Theory, Dover Publications, Inc., 1972 [QA248.S959].
"x is an element of A ∩ B if and only if x is an element of A and x is an element of B."
http://en.wikipedia.org/wiki/Logical_conjunction
"The intersection used in set theory is defined in terms of a logical conjunction: x ∈ A ∩ B if and only if ( x ∈ A ) ∧ ( x ∈ B ). "
http://us.metamath.org/mpegif/elin.html
"Expansion of membership in an intersection of two classes. Theorem 12 of [Suppes] p. 25.
⊢ (A ∈ (B ∩ C) ↔ (A ∈ B ⋀ A ∈ C)) "
----
http://en.wikipedia.org/wiki/Union_(set_theory)
"Formally, x is an element of A ∪ B ∪ C if and only if x is in A or x is in B or x is in C."
http://en.wikipedia.org/wiki/Logical_disjunction
"The union used in set theory is defined in terms of a logical disjunction: x ∈ A ∪ B if and only if ( x ∈ A ) ∨ ( x ∈ B ). "
http://us.metamath.org/mpegif/elun.html
"Expansion of membership in class union. Theorem 12 of [Suppes] p. 25.
⊢ (A ∈ (B ∪ C) ↔ (A ∈ B ⋁ A ∈ C))"
Suppes, Patrick, Axiomatic Set Theory, Dover Publications, Inc., 1972 [QA248.S959].
You contradicted you previous post which said 'or.' Don't go ballistic on my ***.
In fact this is stupid, mathematics is a language other than English.
http://72.3.253.76:8080/webMathematica3/qu...;1/4&variable=x
In fact this is stupid, mathematics is a language other than English.
http://72.3.253.76:8080/webMathematica3/qu...;1/4&variable=x
QUOTE (Beer w/Straw+Mar 4 2009, 07:55 PM)
Would it be the "-13/4<x<-11/4 INTERSECTION x does not equal -3"?
Yes -- -13/4< x < -11/4 AND x != -3
another way to write this is:
-13/4 < x < -3 OR -3 < x < -11/4
Yes -- -13/4< x < -11/4 AND x != -3
CODE
<--- -7/2 --- -13/4 --- -3 --- -11/4 --- -5/2 --->
. **************** -13/4 < x < -11/4
************************* ************************ x != -3
. ******** ******* -13/4 < x < -11/4 AND x != -3
. **************** -13/4 < x < -11/4
************************* ************************ x != -3
. ******** ******* -13/4 < x < -11/4 AND x != -3
another way to write this is:
-13/4 < x < -3 OR -3 < x < -11/4
CODE
<--- -7/2 --- -13/4 --- -3 --- -11/4 --- -5/2 --->
. ******** -13/4 < x < -3
. ******* -3 < x < -11/4
. ******** ******* -13/4 < x < -3 OR -3 < x < -11/4
. ******** -13/4 < x < -3
. ******* -3 < x < -11/4
. ******** ******* -13/4 < x < -3 OR -3 < x < -11/4
Stop it.
Next time I'll ask more mathematical questions that I can't answer.
Next time I'll ask more mathematical questions that I can't answer.
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