# Inequality - confusion

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#1
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!
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2 months ago
#2
(Original post by 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!
what is different in the two cases? Taking reciprocal of positive terms will flip the inequality, but should be equivalent to what you describe?
Last edited by mqb2766; 2 months ago
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#3
(Original post by mqb2766)
what is different in the two cases?
Well, I see there reciprocal of both numbers is taken out,
Last edited by SlowElectron; 2 months ago
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2 months ago
#4
(Original post by 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!
If n is a positive integer, the inequality is never true (except for n=0).

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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2 months ago
#5
(Original post by SlowElectron)
Well, I see there reciprocal of both numbers is taken out,
2 > n+1
results in both cases?
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#6
(Original post by mqb2766)
2 > n+1
results in both cases?
yes and it will result in n < 1 but it should be n > 1.
Last edited by SlowElectron; 2 months ago
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2 months ago
#7
(Original post by SlowElectron)
yes and it will result in n < 1 but it should be n > 1.
If n>1, the original inequality is wrong. Just put n=3 in it.
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#8
(Original post by mqb2766)
If n>1, the original inequality is wrong. Just put n=3 in it.
Ok, but here is another problem.
Logically the result should be
Is it correct? Maybe I'm just unlucky.
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2 months ago
#9
(Original post by SlowElectron)
Ok, but here is another problem.
Logically the result should be
Is it correct? Maybe I'm just unlucky.
-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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#10
(Original post by mqb2766)
-1 satisfies the second one, but not the first. What steps did you do? X being small and negative satisfies the first inequality.
Can you directly tell me what's the correct solution? I can't spend 1 hour on this mistake...
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2 months ago
#11
(Original post by SlowElectron)
Can you directly tell me what's the correct solution? I can't spend 1 hour on this mistake...
Why not just post what you tried?
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#12
(Original post by mqb2766)
Why not just post what you tried?
2x-4 >= 7x
-4 >= 5x
-4/5 >= x
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2 months ago
#13
(Original post by SlowElectron)
2x-4 >= 7x
-4 >= 5x
-4/5 >= x
In the first line, you assume x>0 as you don't flip signs when multiplying both sides by x. This is impossible given the last line.
So assume x<0 when you start and multiply through the inequality.
1
#14
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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2 months ago
#15
(Original post by SlowElectron)
Drawing the graps I see what you mean. But how do you assume x<0? Is there an algebraic method to do that?
Not an algebraic method. You just need to be careful when both sides of an equation are divided by a variable (x in this case).
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2 months ago
#16
If you don't know the sign
* Try both, see which gives consistent solutions
* Sketch the graph

1/x < 1
Last edited by mqb2766; 2 months ago
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