# Inequality - confusion

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#2

(Original post by

Let's say we have the following inequality:

Why do we flip the sign if I'm going to multiply it by and then by ?

Shouldn't it stay the same if I am using a positive integer?

Thanks!

**SlowElectron**)Let's say we have the following inequality:

Why do we flip the sign if I'm going to multiply it by and then by ?

Shouldn't it stay the same if I am using a positive integer?

Thanks!

Last edited by mqb2766; 2 months ago

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(Original post by

what is different in the two cases?

**mqb2766**)what is different in the two cases?

Last edited by SlowElectron; 2 months ago

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#4

**SlowElectron**)

Let's say we have the following inequality:

Why do we flip the sign if I'm going to multiply it by and then by ?

Shouldn't it stay the same if I am using a positive integer?

Thanks!

But if n doesnt have to be an integer:

mutliply both sides by 2(n+1), you get n+1<2 which must be true if your statement holds.

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Last edited by SlowElectron; 2 months ago

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#7

(Original post by

yes and it will result in n < 1 but it should be n > 1.

**SlowElectron**)yes and it will result in n < 1 but it should be n > 1.

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(Original post by

If n>1, the original inequality is wrong. Just put n=3 in it.

**mqb2766**)If n>1, the original inequality is wrong. Just put n=3 in it.

Logically the result should be

Is it correct? Maybe I'm just unlucky.

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#9

(Original post by

Ok, but here is another problem.

Logically the result should be

Is it correct? Maybe I'm just unlucky.

**SlowElectron**)Ok, but here is another problem.

Logically the result should be

Is it correct? Maybe I'm just unlucky.

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(Original post by

-1 satisfies the second one, but not the first. What steps did you do? X being small and negative satisfies the first inequality.

**mqb2766**)-1 satisfies the second one, but not the first. What steps did you do? X being small and negative satisfies the first inequality.

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#11

(Original post by

Can you directly tell me what's the correct solution? I can't spend 1 hour on this mistake...

**SlowElectron**)Can you directly tell me what's the correct solution? I can't spend 1 hour on this mistake...

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#13

So assume x<0 when you start and multiply through the inequality.

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Drawing the graps I see what you mean. But how do you assume x<0? Is there an algebraic method to do that?

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#15

(Original post by

Drawing the graps I see what you mean. But how do you assume x<0? Is there an algebraic method to do that?

**SlowElectron**)Drawing the graps I see what you mean. But how do you assume x<0? Is there an algebraic method to do that?

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#16

If you don't know the sign

* Try both, see which gives consistent solutions

* Sketch the graph

Just think about

1/x < 1

* Try both, see which gives consistent solutions

* Sketch the graph

Just think about

1/x < 1

Last edited by mqb2766; 2 months ago

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