Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

inequalities

I just realised that when you reciprocate both sides of

The inequality sign depends on whether

and

have the same sign or not! Does this now mean that whenever I solve such an equation using reciprocation I need to actually know their signs, or specify the different cases? What a headache!

Example (i made it up):
Frank travels

kilometres north and

. How many kilometres north did he travel? (Note that you can travel a negative number of kilometres north)

So

Similarly, and infinitely more importantly, you cannot multiply both sides by of an equation by

if you don't know its sign! Which means solving ordinary problems like

for

is Impossible! Or... how would you do it?

I used to think you could simply perform normal operations and only flip when u multiply by a negative, but now I realise there could be negative signs in disguise !

Please tell me im wrong somehow lol...

JohnnyReinB

Member

Registered: 2007-10-08

Posts: 453

Re: inequalities

Even if it is negative, you will still flip the sign when you divide by a negative number for example:

-5<4
-5(-2)<(-2)4
10<?-8
10>-8

See?

Just a thought though...

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"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted"

Nisi Quam Primum, Nequequam

JohnnyReinB

Member

Registered: 2007-10-08

Posts: 453

Re: inequalities

Oh, you mean if a is negative or not, and you will divide by it...

I'm guessing you have to solve a first, if not, ask your teacher!

Or, you could assume a is positive, until it is positively proven that it is negative...

I guess you have to do it by trial or error:/

You're right, this is confusing!

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"There is not a difference between an in-law and an outlaw, except maybe that an outlaw is wanted"

Nisi Quam Primum, Nequequam

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: inequalities

(i)
If a and b are both positive or both negative, then 1⁄ab is positive. In this case,

(ii)
If a is negative and b is positive, then 1⁄a is still negative and 1⁄b is still positive. In this case,

(since negative < positive).

In general, if you don’t know know whether a and b have the same sign, you can’t tell how their reciprocals are related. But if you do know, then you will definitely also know how the reciprocals are related.

Last edited by JaneFairfax (2007-11-06 00:29:11)

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: inequalities

For the first problem Jane, aren't you assuming that a is positive? What if you assumed it was negative:

  !!!!!

Even worse, since a is assumed to be negative,

is more correct!!!

It's as if it has a mind of its own!

Last edited by Identity (2007-11-06 00:24:40)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: inequalities

Oops. Sorry, I didn’t realize you said you could travel negative kilometres.

If a is negative, then

is always true, whatever the value of a. (Negative is always less than positive.) That’s all you can say. You can’t do anything else, I’m afraid.

Last edited by JaneFairfax (2007-11-06 00:36:03)

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: inequalities

Thanks. I just realised all the inequalities I've ever done at school involve a being in the numerator, so we don't have to deal with these problems... weird stuff

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: inequalities

As you said, one way of solving such problems would be to split it into the cases when a>0 and a<0. (Presumably when a=0 the problem would become trivial)

However, another way would be to restrict yourself to multiplying by squares of things.
They are always positive and so the inequality sign will be unaffected. The downside is that you might complicate it by introducing higher order terms.

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Why did the vector cross the road?
It wanted to be normal.

Khushboo

Member

Registered: 2007-10-16

Posts: 47

Re: inequalities

a<b

1/a?1/b

a can be a positive or negative number. Say if a is a whole number and when we are reciprocating the number i.e 1/a, we are actually converting the number in a fraction and hence breaking the number in smaller parts. Also larger the denominator smaller is the number, hence if a<b, it means b is greater than a. When we reciprocate both the numbers, b being greater produces a smaller value than a. Thus 1/a>1/b or 1/b<1/a.

Same approach can be used for negative values too. However this is just an observation.

regrads
khushboo

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,219

Re: inequalities

I remember having read this in a Mathematics book of Mir Publishers (I think!)

The Questions is 'What's wrong in this proof?'

Step 1 : 4 < 8

Step 2 : 1/4 > 1/8

Step 3: (1/2)² > (1/2)³

Taking logarithm on both sides,
Step 4: 2 log (1/2) > 3 log (1/2)

Cancelling log(1/2) on both sides,
Step 5: 2>3.

I shall post the answer too here.
When we go from Step 3 to Step 4, that is after step 3, after taking log on both sides, the inequality ought to have changed direction. Because log 1/2 is a negative number (In base 10, log 1/2 is Bar0.6990 or -0.3010 roughly). That is, in an inequality, when both the sides are multiplied by a negative number, the direction of the inequality got to be changed.

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.