Why the Sign Changes from < 0 to > 0 ?

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Why the sign change from < 0 to > 0?
 

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Math_QED
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This just applies the following:

##a>0,b<0\implies ab<0##
 
fresh_42
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This just applies the following:

##a>0,b<0\implies ab<0##
Or ##ab<0 \Longrightarrow (a < 0 \wedge b>0) \vee (a>0 \wedge b<0)##
 
HallsofIvy
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It would have helped if you had told us what the problem was!

IF the problem was to solve the inequality [itex](1- cos(x))(cos(x)- sin(x))< 0[/itex] then
either
a) [itex]`1- cos(x)> 0[/itex] and [itex]cos(x)- sin(x)> 0[/itex]
or
b)[itex]cos(x)- sin(x)> 0[/itex] and [itex]`1- cos(x)> 0[/itex].
That is because "positive times positive is positive" and "negative time negative is negative".

But you seem to have left out the second case.
 
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Thank you for all of your answers but I still don't understand. I even can't find it in my algebra textbook. In what book is this kind of inequality is taught?
 
PeroK
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Thank you for all of your answers but I still don't understand. I even can't find it in my algebra textbook. In what book is this kind of inequality is taught?
In high school I learned that the product of two negative numbers is positive. And the product of a negative and a positive number is negative.

In general, these inequalities will be in any high school maths textbook or syllabus.
 
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Why not like this?

##(1 - \text{cos} ~x)(\text{cos} ~x - \text{sin} ~x) < 0##

##(1 - \text{cos} ~x) > 0 ~\vee~ (\text{cos} ~x - \text{sin} ~x) < 0##
 
fresh_42
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Why not like this?

##(1 - \text{cos} ~x)(\text{cos} ~x - \text{sin} ~x) < 0##

##(1 - \text{cos} ~x) > 0 ~\vee~ (\text{cos} ~x - \text{sin} ~x) < 0##
OR allows only one to be true, so ##(1 - \text{cos} ~x) > 0 ~\vee~ (\text{cos} ~x - \text{sin} ~x) < 0## is true in case ##(1 - \text{cos} ~x) > 0 ~\wedge~ (\text{cos} ~x - \text{sin} ~x) > 0##, whereas ##(1 - \text{cos} ~x)(\text{cos} ~x - \text{sin} ~x) < 0## is false.

##ab<0## means exactly one factor is negative while the other one has to be positive. OR doesn't not gurantee the second condition.
 
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OR allows only one to be true,
No, either or both operands can be true. Possibly you're thinking of "exclusive or" (XOR).
 
fresh_42
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No, either or both operands can be true. Possibly you're thinking of "exclusive or" (XOR).
Yeah, that was a misleading typeset. I didn't meant the Boolean OR, I wanted to emphasize:
<quote> ab < 0 means a>0 and b<0 </quote> or "[ab<0] allows only one to be true"... followed by the explanation that ab<0 is stronger than the Boolean OR. A bit clumsy I admit.
 
FactChecker
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either
a) [itex]`1- cos(x)> 0[/itex] and [itex]cos(x)- sin(x)> 0[/itex]
or
b)[itex]cos(x)- sin(x)> 0[/itex] and [itex]`1- cos(x)> 0[/itex].
You didn't mean that. Those two cases are the same.

@askor , The product of two numbers is negative only if exactly one of the numbers is negative and the other is positive.
That gives two cases:
a) 1-cos x < 0 and (cos x - sin x) >0 As in your original attachment
or
b) 1-cos x > 0 and (cos x - sin x) < 0
The first case, (a), which was in your attachment, is not possible because 1-cos x < 0 implies that 1 < cos x. That does not happen for any real number, x.

That only leaves the second case (b) as a possibility.
So we can say that 1-cos > 0 and (cos x - sin x)<0
 
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PeroK
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PeroK
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I don't know.
Well, that's an honest answer. Can you think of a way to find out?

You could multiply some numbers together and note whether the product is postive or negative. Then you could look for a pattern in the cases where the product is positive.
 
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Well, that's an honest answer. Can you think of a way to find out?

You could multiply some numbers together and note whether the product is postive or negative. Then you could look for a pattern in the cases where the product is positive.
Do you mean if ##ab > 0## then ##(a < 0 \wedge b < 0)## or ##(a > 0 \wedge b > 0)##?
 
PeroK
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Do you mean if ##ab > 0## then ##a < 0 \wedge b < 0## or ##a > 0 \wedge b > 0##?
Personally I'd be happy to say that ##ab > 0## if both ##a## and ##b## are positive or both ##a## and ##b## are negative. Which is what you've written in "wedge" notation.
 
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Personally I'd be happy to say that ##ab > 0## if both ##a## and ##b## are positive or both ##a## and ##b## are negative. Which is what you've written in "wedge" notation.
Do you mean I was correct?
 
PeroK
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Yes. But, if you don't understand that notation properly I wouldn't use it.
The "##\wedge##" notation you mentioned about, it mean "and" isn't it?
 
PeroK
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PeroK
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What is the name of this inequality?
 
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