Rational Numbers
- Thread starter BaileyC
- Start date
it is actually -5 2/3 and -4 2/3,
It is a mixed fraction
seriously get it right and all of you people maybe instead of critising by lack of math skills you could... oh... i don't know? HELP ME! Maybe instead you could do that, it is the whole purpose of thhis site.
Just sayin'' :evil:
It is a mixed fraction
seriously get it right and all of you people maybe instead of critising by lack of math skills you could... oh... i don't know? HELP ME! Maybe instead you could do that, it is the whole purpose of thhis site.
Just sayin'' :evil:
- Joined
- Apr 12, 2005
- Messages
- 11,339
There is no criticizing going on here. Just answer the question.
"It is actually -5 2/3 and -4 2/3,"
You're kidding, right? I can list nine (9) as quickly as I can type.
Please show us one!
Really, what is the definition of a Rational Number? Do you have it? Something about "quotient" and "Integer".
I'd start this way: -5 2/3 = -17/3 and -4 2/3 = -14/3
In this case, it should be mortally obvious that -15/3 and -16/3 are such rational numbers.
Note, that -15/3 = -5
You tell me a third rational number in the requested range.
"It is actually -5 2/3 and -4 2/3,"
You're kidding, right? I can list nine (9) as quickly as I can type.
Please show us one!
Really, what is the definition of a Rational Number? Do you have it? Something about "quotient" and "Integer".
I'd start this way: -5 2/3 = -17/3 and -4 2/3 = -14/3
In this case, it should be mortally obvious that -15/3 and -16/3 are such rational numbers.
Note, that -15/3 = -5
You tell me a third rational number in the requested range.
BaileyC said:
BaileyC said:
I sent a message to Ted regarding the bad behavior, bad attitude, and misrepresentation
by user BaileyC from the posts above.
In no way are those fractions, you originally gave, mixed numbers, BaileyC.
They are, without ambiguity, improper fractions.
*You* made the errors in showing your intended numbers the wrong way,
and then *you* blamed your errors on us (you: "seriously get it right").
You are not worthy of any help whatsoever until you admit that you made the errors,
not us, and you come to the forum with *humility*.
It is *not* "the whole purpose of this site." The purpose of this site is to critique and
possibly give helpful suggestions to users who show attempts at their work here,
including not misrepresenting what they have originally typed, and who are not
disrespectful to the people giving the assistance, as you have been doing.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
BaileyC said:
I cannot understand what you're trying to ask.
My best suggestion is that you speak with your teaching about help obtaining a face-to-face tutor.
You're in need of intensive help, and we cannot provide that level over the Internet.
Bailey, it boils down to you not knowing even enough to be able to ask a proper question.
That's due to YOU having a bad teacher, or to YOU skipping math classes...
How would you feel if someone asked you to teach him how to ride a bicycle,
but this guy had no idea what a bicycle is?
That's due to YOU having a bad teacher, or to YOU skipping math classes...
How would you feel if someone asked you to teach him how to ride a bicycle,
but this guy had no idea what a bicycle is?
J
JeffM
Guest
BaileyC said:Uhh... Ok yea, I used bad attitude... whatevs we can move on right
BUT I HAVE ONE QUESTION??? are the improper fractions made rational numbers or the number made out of the improper fraction. Sorry? I am just struggling with math.
Yes, you did exhibit a bad attitude in not understanding that people here are doing favors for people. Being polite is a good habit. But, as you say, let's move on.
Second, you will get better help by being as careful in your questions as you can be.
Third, doing well in math is largely a matter of making sure you understand concepts fully and then practicing with lots of examples to cement the concept in your own mind.
Fourth, speaking a bit too casually, a rational number is one that can be expressed as a proper or improper fraction involving only integers. That is, the numerator is any integer, and the denominator is any integer other than zero. That is the concept. Now go back to your original question and see if, GIVEN THAT CONCEPT, you can come up with a few examples other than the ones already given to you so that you can prove to yourself that you have mastered the concept. Bring your answer here, and I at least will critique it. (By the way, there are an infinite number of correct answers so the process should not take long.)
Hint (1): Is -18 greater or less than - 17?
Hint (2): Is - (17 / 3) equivalent to - (170 / 30)?
Hello, BaileyC!
\(\displaystyle \text{We have: }\:\begin{Bmatrix}\text{-}5\frac{2}{3} &=& \text{-}\frac{17}{3} &=& \text{-}\frac{34}{6} \\ \\[-3mm] \text{-}4\frac{2}{3} &=& \text{-}\frac{14}{3} &=& \text{-}\frac{28}{6}\end{Bmatrix}\)
\(\displaystyle \text{Can we find three rational numbers between }\text{-}\tfrac{34}{6}\text{ and }\text{-}\tfrac{28}{6}\,?\)
\(\displaystyle \text{Sure! \;Take your pick:}\)
. . \(\displaystyle \left\{\text{-}\tfrac{33}{6},\;\text{-}\tfrac{32}{6},\;\text{-}\tfrac{31}{6},\;\text{-}\tfrac{30}{6},\;\text{-}\tfrac{29}{6}\right\} \;=\;\left\{\text{-}5\tfrac{1}{2},\;\text{-}5\tfrac{1}{3},\;\text{-}5\tfrac{1}{6},\;\text{-}5,\;\text{-}4\tfrac{5}{6} \right\}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
\(\displaystyle \text{Another approach . . .}\)
\(\displaystyle \text{We have: }\:\begin{Bmatrix}\text{-}5\frac{2}{3} &=& \text{-}\frac{17}{3} \\ \\[-3mm] \text{-}4\frac{2}{3} &=& \text{-}\frac{14}{3} \end{Bmatrix}\)
\(\displaystyle \text{We see two numbers immediately: }\:x = \text{-}\tfrac{16}{3}\text{ and }y = \text{-}\tfrac{15}{3}\)
\(\displaystyle \text{Can we find a third number? . . . Yes!}\)
\(\displaystyle \text{Take the }average\text{ of }x\text{ and }y.\)
. . \(\displaystyle \text{It lies exactly halfway between }x\text{ and }y.\)
\(\displaystyle \text{So we have: }\:\frac{(\text{-}\frac{16}{3}) +(\text{-}\frac{15}{3})}{2} \:=\:\frac{\text{-}\frac{31}{3}}{2} \:=\:\text{-}\frac{31}{6}\)
\(\displaystyle \text{And the three numbers are: }\:\left\{\text{-}\tfrac{16}{3},\;\text{-}\tfrac{31}{6},\;\text{-}\tfrac{15}{3}\right\} \;=\;\left\{\text{-}5\tfrac{1}{3},\;\text{-}5\tfrac{1}{6},\;\text{-}5\right\}\)
\(\displaystyle \text{We have: }\:\begin{Bmatrix}\text{-}5\frac{2}{3} &=& \text{-}\frac{17}{3} &=& \text{-}\frac{34}{6} \\ \\[-3mm] \text{-}4\frac{2}{3} &=& \text{-}\frac{14}{3} &=& \text{-}\frac{28}{6}\end{Bmatrix}\)
\(\displaystyle \text{Can we find three rational numbers between }\text{-}\tfrac{34}{6}\text{ and }\text{-}\tfrac{28}{6}\,?\)
\(\displaystyle \text{Sure! \;Take your pick:}\)
. . \(\displaystyle \left\{\text{-}\tfrac{33}{6},\;\text{-}\tfrac{32}{6},\;\text{-}\tfrac{31}{6},\;\text{-}\tfrac{30}{6},\;\text{-}\tfrac{29}{6}\right\} \;=\;\left\{\text{-}5\tfrac{1}{2},\;\text{-}5\tfrac{1}{3},\;\text{-}5\tfrac{1}{6},\;\text{-}5,\;\text{-}4\tfrac{5}{6} \right\}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
\(\displaystyle \text{Another approach . . .}\)
\(\displaystyle \text{We have: }\:\begin{Bmatrix}\text{-}5\frac{2}{3} &=& \text{-}\frac{17}{3} \\ \\[-3mm] \text{-}4\frac{2}{3} &=& \text{-}\frac{14}{3} \end{Bmatrix}\)
\(\displaystyle \text{We see two numbers immediately: }\:x = \text{-}\tfrac{16}{3}\text{ and }y = \text{-}\tfrac{15}{3}\)
\(\displaystyle \text{Can we find a third number? . . . Yes!}\)
\(\displaystyle \text{Take the }average\text{ of }x\text{ and }y.\)
. . \(\displaystyle \text{It lies exactly halfway between }x\text{ and }y.\)
\(\displaystyle \text{So we have: }\:\frac{(\text{-}\frac{16}{3}) +(\text{-}\frac{15}{3})}{2} \:=\:\frac{\text{-}\frac{31}{3}}{2} \:=\:\text{-}\frac{31}{6}\)
\(\displaystyle \text{And the three numbers are: }\:\left\{\text{-}\tfrac{16}{3},\;\text{-}\tfrac{31}{6},\;\text{-}\tfrac{15}{3}\right\} \;=\;\left\{\text{-}5\tfrac{1}{3},\;\text{-}5\tfrac{1}{6},\;\text{-}5\right\}\)
D
Deleted member 4993
Guest
tkhunny said:
Why should s/he - now that the answer, the whole answer and everything about the whole answer has been given - why would s/he lift the pencil but to copy the answers??