inequality, absolute value, etc.

[PG1995]

Hi

Could you please help me with this query? Thank you.

Regards
PG

 

[MrAl]

Hi,

Remember way back when i told you that when you multiply by the variable you change the solutions. You can get away with that if you are careful, but i think there is a better way.
Multiplication and division of the variable is tricky, so we'll stick to addition and subtraction for now.

Starting with:
(2*x-5)/(x-2)<1

subtract 1 from both sides and we get:
(2*x-5)/(x-2)-1<0

and simplifying the left side we get:
(x-3)/(x-2)<0

Now set the numerator equal to zero and solve for x we get:
x=3

then set the denominator equal to zero and solve for x and we get:
x=2

What this gives us is two critical points. The true solutions can lie between these two, to the right of 3, or to the left of 2, or any two of those intervals or all three. So we have to check a point in each interval to see if it is a valid interval, and we also of course have to check the critical points themselves to make sure they work and do not generate an infinity.

So the result is:
2<x<3

Also realize that some results in other problems may come out with the middle interval missing so we would not be able to state the result so neatly but have to resort to notation like:
{-inf to 2} or {3 to +inf}
or something like that (maybe using brackets).
This problem was neat in that the intervals were all connected or that only one interval was valid.

Also, some problems may come out with one of the intervals that happens to contain 0, but 0 isnt really part of the solution, so zero has to be checked for validity as well.

 

[PG1995]

Thank you, MrAl.

Actually the way you have guided me to solve it, using addition and subtraction, is exactly the same as it is solved in the book. So, I'm interested in finding the solution using the other method where we multiply by (x-2). As you have already seen that my attempt with this method was futile. Could you please help me with it? Thanks.

Regards
PG

 

[MrAl]

Hi,

Wow that's amazing. It's almost the same except they took the time to graph the solution too, which was nice. I figured you could do that if you wanted to. They also show the open points due to testing of the critical points.

But why would you want to make life difficult? What reason do you have for wanting to solve it another way?

Another way is to simply graph the function itself and look for the critical points. To make them more apparent if you have a program that can graph for you and it handles inequalities as logical statements then you can graph the entire function. That will be a square wave with height equal to 1 when the function passes and height equal to 0 when it fails. The leading and trailing edges of this new logic function will show you the critical points and automatically test the intervals.
If the graph program can not handle inequalities as logic statements, then you might be able to graph a statement like this:
if F(x)<1 then 1 else 0
and that will do the same thing. F(x) must be the original function.

Remember again though that when you multiply or divide by the variable you change the information so you may change one or more solutions, so you may never get the right answer that way without going back and considering the original expression anyway. I could probably come up with a better example of this too, but in your original attempt your second function failed already didnt it, and that is what happens. So if you try it that way you'll just have to test the other functions too and make sure you didnt miss anything.

So i repeat my question, why would you want to try to do it the other way?

 

[PG1995]

Once again, thanks, MrAl.

But why would you want to make life difficult? What reason do you have for wanting to solve it another way?

I believe a simple reason would be curiosity. But I think it's better to use the technique of 'addition and subtraction' than going for some difficult and error prone method. So, we can stop this discussion here about that query. Perhaps, some other day we can look at that other method more carefully. Thank you.

Regards
PG

 

[PG1995]

Hi

Could you please help me with this query? Thank you.

Regards
PG

 

[MrAl]

Hi,

In short, for an inequality such as:
a<=b

which we read "a is less than or equal to b"

can be an expression or a partially self fulfilling standing statement. That is because the inequality is "less than OR equal to", so it does not have to be both because of the "OR" in there. You might be reading it as:

"less than AND equal to"

which would not be the same thing. So for the statement:
0<=2

for this to be true zero must be less than or equal to 2. It is not equal to 2 but it is less than 2 so it works ok. Another true statement is:
1<=2

and also is:
2<=2

but this time it's the equal sign that gives it credence. On the other hand, if we see:

3<=2

then that is false because 3 is not less than 2 and it is also not equal to 2.

 

[steveB]

PG,

Perhaps what is bothering you about those examples is that they use numbers, rather than variables. Although, the expressions are logically correct, they might seems silly when the numbers are known. You may wonder, "why include the false part, and why not only retain the true part, and why use the or-operation at all". Well, in practice you rarely see those types of expressions with constant known numbers, but consider the use of variables. In programming we might write an if-statement as "if x>=2 then ... ". Since we now don't know what the x-value is beforehand, you might feel more comfortable with this. Sometimes x<2, which makes the inequality false, and sometimes x=2 which makes the inequality true and sometimes x>2 which also makes the inequality true.

Steve

 

[PG1995]

Thank you, MrAl, Steve.

You might be reading it as: "less than AND equal to" which would not be the same thing.

No, I wasn't reading it with "and". But your examples really helped me. Thanks.

Although, the expressions are logically correct, they might seems silly when the numbers are known. You may wonder, "why include the false part, and why not only retain the true part, and why use the or-operation at all".

Exactly. It really seems silly when the numbers are known and this is what troubles me. I don't have any problem when variables are used.

Regards
PG

 

[steveB]

So, if mathematicians cared about the "silly-bit" being "on" or "off", then this would matter, but it seems they only care whether the logic-bit is "true" or "false".

-joking

 

[MrAl]

Thank you, MrAl, Steve.



No, I wasn't reading it with "and". But your examples really helped me. Thanks.



Exactly. It really seems silly when the numbers are known and this is what troubles me. I don't have any problem when variables are used.

Regards
PG

Hi,

Well to add a little here, when they state something like 1<=2 they are not just stating that 1 is less than or equal to 2, they are stating that the statement "1<=2" is "True", not "False". So there is really more information to look at when trying to understand what they mean by 1<=2. It's not that they are making a statement about facts relating to ONLY 1 and 2, they are making a statement about facts relating to 1, 2, and also "true" (or "false").
So you can not read it with regard to just 1 and 2, you also have to read it with regard to true or false.

Lets look at a quick example...

The number of cars in lot A (nA) is less than or equal to the number of cars in lot B (nB). So we would have:
nA<=nB

Now does this mean that we can never have less cars parked in lot A than in lot B? Of course not. But it does tell us that the number of cars in lot B will always be more than the number in lot A or else the numbers will be equal.

Another way to look at this when the numbers are integers such as cars, is:
nA<(nB+1)

There you have a different relationship that says the same thing without using the equals sign.

So in the future we count the cars in both lots. In lot A we find 9 cars and in lot B we find 22 cars, so we want to see if our relationship nA<=nB holds true so we use the test:
9<=22

which means:
If either 9 is less than 22 or 9 is equal to 22 then the statement is true, but if not then it is false.
Since 9 is less than 22 it meets one of the criteria that makes the statement true, so it's true.

But the 9<=22 does not stand alone all by itself because there is more meaning attached to it and that it whether or not it is true or false, and in the life situation there are less cars in lot A than B so the "Law of cars in lots A and B" holds true.

And again, for integers we can change the statement to:
nA<(nB+1)

which for our example of 9 and 22 would then convert to:
9<22+1

or simply:
9<23

but for all numbers we would have to do something like:
nA<(nB+dx)

where dx is some very very infinitesimally small number ideally 1/infinity.

 

[PG1995]

Thank you, MrAl.

I don't get the point where you say the following.

Another way to look at this when the numbers are integers such as cars, is:
nA<(nB+1)

There you have a different relationship that says the same thing without using the equals sign.

Could you please elaborate on it a little? Why do you have (nB+1)? I think due to the addition of "1" to nB the inequality, nA<(nB+1), can never be equal because nA is always to be less than (nB+1). Thanks.

Regards
PG

 

[MrAl]

Hello again PG,


Well, that's because that equivalent statement is NEVER supposed to have the left side equal to the right side. In the original statement that was a possibility, but if the new statement ever becomes equal then it becomes false and that is necessary when we increment the right hand side alone.

The statement:
nA<=nB

is exactly equivalent to the statement:
nA<(nB+1)

when we are working with integers, but we were able to eliminate the equals sign with that trick.

Lets look at few examples and i think it will be more clear. We'll compare the old statement to the new statement for the same range of integers. I'll have to do this in code so that it comes out aligned for more easy reading.

We will begin by making nB a constant equal to 2 and vary nA from 0 to 4:

nA<=nB    nA<(nB+1)
 0<=2       0<3        Both true
 1<=2       1<3        Both true
 2<=2       2<3        Both true
 3<=2       3<3        Both false
 4<=2       4<3        Both false

Notice now that when we have the case 3<=2 that is false, and the new statement reads 3<3 which
is also false. Where the transition from true to false occurs in the old statement obviously
3 is greater than 2 so it cant be true, while in the new statement 3 is equal to 3 so it can
not be said that 3 is less than 3 so that is false also. Wherever the original statement is true
the new statement is true, and wherever the old statement is false the new statement is also false.
So if you were to graph this you would get the same results for both functions.
Also you must remember that we drop the equals sign for the new statement as that was the main reason for
doing this in the first place.

 

[PG1995]

Thank you, MrAl.

I understand it now.

Regards
PG

 

[PG1995]

Hi

I myself don't exactly know what troubles me in this case, Q1, but I have stated my own explanation which I'm not really satisfied with. Perhaps, you can guide me in proper direction. Thanks.

Could you please also help me with these queries, Q2 and Q3(a) and Q3(b)?

Regards
PG

 

[MrAl]

Hi,

Not sure what your question is for the inequalities this time.

Real quick:
The solution to the cubic has a repeated root at x=1. To prove this, you should be able to multiply all the x minus each root and get the same equation as such:
(x-r1)*(x-r2)*(x-r3)*...*(x-rN)=same equation

So for this example we have:
(x-1)*(x-1)*(x+2)=x^3-3*x+2

because we had a repeated root of +1 and a single root of -2.

Note this works because if the original equation was equal to zero:
x^3-3*x+2=0

then the 'new' equation knowing the negated roots must be equal to zero:
(x-1)*(x-1)*(x+2)=0

and that means that each product must be equal to zero:
(x-1)=0
(x-1)=0
(x+2)=0

Or you could look at the other way around: because each product is zero the new equation is zero and so the expanded version must be the same as the original.

 

[steveB]

PG,

For the inequality proofs, I recommend you proceed along the same lines that the book uses. Operations with the absolute value operator tend to be awkward, which, I'm guessing, is why you don't like the proofs in the book. Perhaps there is a more elegant way (maybe using geometry arguments would be a way to explore) but if you just work through all permutations of "a" and "b" that let you get rid of the absolute value operators, it should work out on a case by case basis.

A literature search (maybe even an internet search) might uncover some more elegant approaches, which would no doubt be interesting to see.

For example, if a and b are positive and a>b, then for one relation you get a-b<=a+b, which is clearly true, and for the other relation you get a-b<=a-b which is clearly true. OK, I did the easy one, but the others will not be hard. Perhaps the hard part is identifying all the relevant cases, and proving you didn't miss any.

Steve

 

[steveB]

PG,

On your Q1, it looks like the only thing you missed was the point x=1.5 which is not allowed as a solution according to the book solution. Other than that, it seems you got the same answer correctly.

 

[steveB]

PG,

By the way, I'm a little uncertain about the exclusion of x=1.5, as the book does. To me, the left hand side is not indeterminate, but is a 1/0 type of limit. Since the limit is +∞ from both the left and from the right, I don't see why we can't say that +∞>5 for this problem. Then again, I'm not a mathematician and I think like an engineer. So, I'm not going to argue that case too strongly. Just something to think about.

MrAl,

I'm curious what you think about the exclusion of x=1.5?