Help in proving this inequality using the concepts or theorems on the properties of real numbers

- blackbird02

Help in proving this inequality using the concepts or theorems on the properties of real numbers

- jamiebookeater

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- blackbird02

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- blackbird02

Prove that if a>0, b<0, then \[ab+\frac{ b }{ a }<0\]

- triciaal

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- blackbird02

what does ve stand for?

- triciaal

sorry, -ve = negative and +ve = positive

- blackbird02

I'm sorry, but I still don't get it. What would be my starting equation in the proving?

- zzr0ck3r

\(ab<0 \text{ and } \frac{a}{b}<0\)
so
\(ab+\frac{a}{b}<0\)

- blackbird02

@zzr0ck3r How would I prove this using the concepts or theorems on the properties of real numbers?

- blackbird02

@Hero any idea how to prove this?

- Hero

Are you still here.

- blackbird02

Yeah

- Hero

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- blackbird02

@Hero just a clarification, in the seventh row, what is the equation? is it ab<0 +b/a<0
Or b/a<0 should be on another line?

- Hero

It's one line: Two inequalities being added together

- blackbird02

Oh, okay. I get it now. Thank you so much!

- Hero

It's written exactly as it Should be

- triciaal

@hero can you explain the step where you have "flip sign"?

- Hero

b is negative If you multiply both sides of a>0 by a negative number, you have to invert the inequality symbol

- blackbird02

@Hero Thank you so much!

- triciaal

sorry, well thanks but that's the only place we differ on why a negative times a positive is negative. multiplication is group addition and we have a constant times a negative entity so it is more negative.

- zzr0ck3r

"multiplication is group addition" ?

- triciaal

@zzr0ck3r yes example 3 * 2 = 2 + 2 + 2 etc

- triciaal

maybe repeated addition is better

- zzr0ck3r

addition is the operation on the reals as a group
multiplication is the other operation on the reals as a ring

- zzr0ck3r

how would you write out (1/2)*(1/3) in those terms?

- triciaal

what?

- zzr0ck3r

google groups and rings.

- zzr0ck3r

we start with a group an one operation addition
we extend this idea to another operation called multiplication, once we have two operations and a few more properties we have a ring.
we then include multiplicative inverses to close it up, and we get a field.
All I am saying, is that sentence made no sense. ...

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