# Why the Sign Changes from < 0 to > 0 ?

• B

## Main Question or Discussion Point

Why the sign change from < 0 to > 0?

#### Attachments

• 2.6 KB Views: 59

Math_QED
Homework Helper
2019 Award
This just applies the following:

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

fresh_42
Mentor
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
Homework Helper
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.

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
Homework Helper
Gold Member
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.

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
Mentor
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
Mentor
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
Mentor
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
Gold Member
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

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

PeroK
Homework Helper
Gold Member
Does above applies too when ab > 0?
What are the conditions on $a$ and $b$ if $ab > 0$?

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

PeroK
Homework Helper
Gold Member
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.

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
Homework Helper
Gold Member
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.

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

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

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

PeroK
Homework Helper
Gold Member
Which one is "or"?
$\vee$