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Math_QED

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This just applies the following:

##a>0,b<0\implies ab<0##

##a>0,b<0\implies ab<0##

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Or ##ab<0 \Longrightarrow (a < 0 \wedge b>0) \vee (a>0 \wedge b<0)##This just applies the following:

##a>0,b<0\implies ab<0##

HallsofIvy

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IF the problem was to solve the inequality [itex](1- cos(x))(cos(x)- sin(x))< 0[/itex] then

a) [itex]`1- cos(x)> 0[/itex] and [itex]cos(x)- sin(x)> 0[/itex]

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

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

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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.

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

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

##ab<0## means

Mark44

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No, either or both operands can be true. Possibly you're thinking of "exclusive or" (XOR).OR allows only one to be true,

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Yeah, that was a misleading typeset. I didn't meant the Boolean OR, I wanted to emphasize:No, either or both operands can be true. Possibly you're thinking of "exclusive or" (XOR).

<quote> ab < 0 means a>0 and b<0 </quote>

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You didn't mean that. Those two cases are the same.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].

@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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Does above applies too when ab > 0?Or ##ab<0 \Longrightarrow (a < 0 \wedge b>0) \vee (a>0 \wedge b<0)##

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What are the conditions on ##a## and ##b## if ##ab > 0##?Does above applies too when ab > 0?

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I don't know.What are the conditions on ##a## and ##b## if ##ab > 0##?

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Well, that's an honest answer. Can you think of a way to find out?I don't know.

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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Do you mean if ##ab > 0## then ##(a < 0 \wedge b < 0)## or ##(a > 0 \wedge b > 0)##?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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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 if ##ab > 0## then ##a < 0 \wedge b < 0## or ##a > 0 \wedge b > 0##?

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Do you mean I was correct?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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Yes. But, if you don't understand that notation properly I wouldn't use it.Do you mean it was correct?

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The "##\wedge##" notation you mentioned about, it mean "and" isn't it?Yes. But, if you don't understand that notation properly I wouldn't use it.

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Yes. And the other one means "or".The "##\wedge##" notation you mentioned about, it mean "and" isn't it?

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Which one is "or"?Yes. And the other one means "or".

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What is the name of this inequality?

Mark44

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As far as I know, it's too simple to have a name.What is the name of this inequality?

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