Why the Sign Changes from < 0 to > 0 ?

[askor]

Why the sign change from < 0 to > 0?

 

[member 587159]

This just applies the following:

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

 

[fresh_42]

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]

It would have helped if you had told us what the problem was!

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

But you seem to have left out the second case.

 

[askor]

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]

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.

 

[askor]

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]

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.

 

[Mark44]

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]

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]

either
a) $`1- cos(x)> 0$ and $cos(x)- sin(x)> 0$
or
b)$cos(x)- sin(x)> 0$ and $`1- cos(x)> 0$.

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

 

[askor]

Or $ab<0 \Longrightarrow (a < 0 \wedge b>0) \vee (a>0 \wedge b<0)$

Does above applies too when ab > 0?

 

[PeroK]

Does above applies too when ab > 0?

What are the conditions on $a$ and $b$ if $ab > 0$?

 

[askor]

What are the conditions on $a$ and $b$ if $ab > 0$?

I don't know.

 

[PeroK]

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.

 

[askor]

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]

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.

 

[askor]

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]

Do you mean it was correct?

Yes. But, if you don't understand that notation properly I wouldn't use it.

 

[askor]

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]

The "$\wedge$" notation you mentioned about, it mean "and" isn't it?

Yes. And the other one means "or".

 

[askor]

Yes. And the other one means "or".

Which one is "or"?

 

[PeroK]

Which one is "or"?

$\vee$

That's your last question.

 

[askor]

What is the name of this inequality?

 

[Mark44]

What is the name of this inequality?

As far as I know, it's too simple to have a name.