# What is the necessary and sufficient condition?

• Helly123
In summary: In the sense of arithmetic mean, ##2## is the midpoint of the interval ##(5,-2)##. But we are dealing here with ##x^2 - 3x -10 < 0##, which implies that ##(x - 5)(x + 2) < 0##. So obviously, ##x = 2## is not the midpoint of this interval.In the sense of arithmetic mean, ##2## is the midpoint of the interval ##(5,-2)##. But we are dealing here with ##x^2 -
Helly123

## Homework Statement

|x-2| < a is a necessary condition for x^2 -3x -10 < 10 . What is the range value of a?
|x-2| < a is a sufficient condition for x^2 -3x -10 < 10. What is the range value of a?[/B]
The options are
a>= 4
a>=3
0<a<=2
0<a<=3
0<a<=5

## The Attempt at a Solution

Range of x for x^2 -3x -10 < 0
2<x<5

And
|x-2| < a
Range of a is 0 < a < 3

Necessary = 0<a<=3
What is the necessary and sufficient conditions?

Last edited:
Helly123 said:

## Homework Statement

|x-2| < a is a necessary condition for x^2 -3x -10 < 10 . What is the range value of a?
|x-2| < a is a sufficient condition for x^2 -3x -10 < 10. What is the range value of a?[/B]
The options are
a>= 4
a>=3
0<a<=2
0<a<=3

## The Attempt at a Solution

Range of x for x^2 -3x -10 < 10
2<x<5

And
|x-2| < a
Range of a is 0 < a < 3

Necessary = 0<a<=3
What is the necessary and sufficient conditions?

Your x-range is incorrect: the range of x for ##x^2 - 3x - 10 < 10## is not ##2 < x < 5##.

Are you sure you copied the problem correctly?

Ray Vickson said:
Your x-range is incorrect: the range of x for ##x^2 - 3x - 10 < 10## is not ##2 < x < 5##.

Are you sure you copied the problem correctly?
Sorry. I have corrected it

Helly123 said:
Sorry. I have corrected it

So, where is the correction? All I see is the original problem, not corrected.

Ray Vickson said:
So, where is the correction? All I see is the original problem, not corrected.
Sorry. I forgot to correct the up side of question. The equation is x^2 - 3x -10 < 0

Helly123 said:
Sorry. I forgot to correct the up side of question. The equation is x^2 - 3x -10 < 0

It is certainly possible to give ##|x-2| < a## as a sufficient condition, and there are many values of ##a## that will work; however, there is no value of ##a## that will have ##|x-2| < a## as a necessary condition.

Helly123 said:
Sorry. I forgot to correct the up side of question. The equation is x^2 - 3x -10 < 0
You have not solved this correctly.

In the OP, when you make statement about "Range" I assume that you mean "solution".
Helly123 said:

## The Attempt at a Solution

Range of x for x2 -3x -10 < 0
2<x<5
The solution to ##\ x^2 -3x -10 < 0\ ## is not ##\ 2<x<5\ ##.

For example x = 1 is also a solution to ##\ x^2 -3x -10 < 0\ ##.

By the way, one of the given choices does give a necessary condition.

If ##P\implies Q ##, then we say ##P ## is a sufficient condition for ## Q## and ##Q## is a necessary condition for ## P##
The first problem is to find ##a ## such that
$$x^2 -3x-10<0\implies \lvert x-2\rvert <a$$
One may note that ##x^2 - 3x -10 = (x-5)(x+2)<0 ##, which is true for every ##x ## satisfying ##-2<x<5 ##.
Now examine a condition ##\lvert x-2\rvert <a ## which is the same as
$$-a < x-2 < a \iff 2-a <x<2+a$$
Now, for which largest possible ##a ## do you not exceed the bounds? (NOT a necessary condition, however!)

Conversely, consider the reversed implication
$$\lvert x-2\rvert <a \implies x^2 -3x-10<0$$
Only now, your premise is the given condition ##2-a<x<2+a ##. You must guarantee ##(x+2)(x-5)<0 ##. Which values of ##a## guarantee that?There is a problem with terminology, though. You can only find sufficient conditions of the form ##2-a<x<2+a ##. There are no necessary conditions.
To quickly illustrate what I mean. If ##Q ## is a necessary condition for ## P##, then ##\neg Q## is a sufficient condition for ## \neg P##. In other words, if we did, indeed have a necessary condition ##2-a<x<2+a ## for some ##a##, then ##x\leq 2-a\lor x\geq 2+a ## must be a sufficient condition for ##(x+2)(x-5)\geq 0 ##, but that is not true of any ## a##, therefore a necessary condition of the given form cannot exist.

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Helly123
Not only is there a sufficient condition given, but there is also a necessary condition given.

It's rather futile to discuss either without the OP giving the correct solution to the given quadratic inequality.

##
\ x^2 -3x -10 < 0\ ##​

SammyS said:
Not only is there a sufficient condition given, but there is also a necessary condition given.

It's rather futile to discuss either without the OP giving the correct solution to the given quadratic inequality.

##
\ x^2 -3x -10 < 0\ ##​

NO. If you look at the correct interval ##\alpha < x < \beta## for ##x^2 - 3x - 10 < 0## it just cannot be written as ##|x - 2 | < a## for some ##a##. Basically, ##2## is not the mid-point of the interval ##(\alpha, \beta)##.

Ray Vickson said:
NO. If you look at the correct interval ##\alpha < x < \beta## for ##x^2 - 3x - 10 < 0## it just cannot be written as ##|x - 2 | < a## for some ##a##. Basically, ##2## is not the mid-point of the interval ##(\alpha, \beta)##.
It doesn't need to be centered on any particular point..

I think it best to wait for OP to correctly solve the given quadratic inequality before further discussing this point.

SammyS said:
It doesn't need to be centered on any particular point..

I think it best to wait for OP to correctly solve the given quadratic inequality before further discussing this point.

I agree. Certainly a sufficient condition can be written as ##|x-2| < a##---and that for lots of different values of ##a##. And, of course, if we let ##a = +\infty## we can write a necessary condition ##x \in \mathbb{R}## as ##|x-2| < a.## I don't think that is what the questioner meant, although I could be wrong.

Sorry again.. the value of x for x^2 -3x -10<0 is -2<x<5

When -2<x<2
a value is 0<a<4

When 2<=x<5
0<=a<3

But i cannot decide what is the sufficient and necessary conditions

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Helly123 said:
Sorry again.. the value for x is −2 < x < 5
The above is the correct solution to the given quadratic inequality.

The following makes little to no sense.
For −2<x<2
0<a<4
If a = 2, then |x − 2| < a is equivalent to 0 < x < 4.

For 2<=x<5
0<=a<3

But i cannot decide what is the sufficient and necessary conditions

SammyS said:
The above is the correct solution to the given quadratic inequality.

The following makes little to no sense.

If a = 2, then |x − 2| < a is equivalent to 0 < x < 4.
I have corrected it

Ray Vickson said:
And, of course, if we let ##a = +\infty## we can write a necessary condition ##x \in \mathbb{R}## as ##|x-2| < a.##
That's basically saying if ##f(x)< 0##, then ## x\in\mathbb R##, a vacuously true statement. That said, I doubt they operate in ##\bar{\mathbb R} ##

nuuskur said:
That's basically saying if ##f(x)< 0##, then ## x\in\mathbb R##, a vacuously true statement. That said, I doubt they operate in ##\bar{\mathbb R} ##
Well, of course it is vacuous---that is exactly the point! Anyway, ##\bar{\mathbb{R}}## is not involved, just ##\mathbb{R}##.

##a=+\infty ## is meaningless in that instance, if one insists on being overly pedantic

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nuuskur said:
##a=+\infty ## is meaningless in that instance, if one insists on being overly pedantic

No: it is just notation. Saying that ##|x-2| < \infty## is just saying that ##|x-2|## can be any real number. Personally, I would never voluntarily say it that way, but it would certainly be a way to put the statement in the format requested by the problem.

Helly123 said:
Sorry again.. the value of x for x^2 -3x -10<0 is -2<x<5

When -2<x<2
a value is 0<a<4

When 2<=x<5
0<=a<3

But i cannot decide what is the sufficient and necessary conditions

Yes.
For | x − 2 | < a to be a sufficient condition so that x2 − 3x − 10 < 0, the range for a is 0 < a ≤ 3.
You have not been very clear in showing that this is the case.

This does not give all of the values of x which satisfy the given quadratic inequality. Rather any x fulfilling | x − 2 | < a will satisfy the quadratic inequality.for the specified range of a values. So being in that interval is a sufficient condition for x to satisfy the quadratic inequality..

Helly123
SammyS said:
Yes.
For | x − 2 | < a to be a sufficient condition so that x2 − 3x − 10 < 0, the range for a is 0 < a ≤ 3.
You have not been very clear in showing that this is the case.

This does not give all of the values of x which satisfy the given quadratic inequality. Rather any x fulfilling | x − 2 | < a will satisfy the quadratic inequality.for the specified range of a values. So being in that interval is a sufficient condition for x to satisfy the quadratic inequality..
I don't get it why is 0 < a ≤ 3? Why including 3?

Helly123 said:
I don't get it why is 0 < a ≤ 3? Why including 3?
You beat me to that question.

For one thing, if you look at the problem statement you gave, that choice includes the 3.
Helly123 said:
The options are
a>= 4
a>=3
0<a<=2
0<a<=3
0<a<=5
As a teacher I would be inclined to ask you , "Why does that include the 3?"

Plug a = 3 into the inequality |x−2| < a and solve for x.
What do you get?

Then we can tackle the "necessary" condition.

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SammyS said:
You beat me to that question.

For one thing, if you look at the problem statement you gave, that choice includes the 3.

As a teacher I would be inclined to ask you , "Why does that include the 3?"

Plug a = 3 into the inequality |x−2| < a and solve for x.
What do you get?

Then we can tackle the "necessary" condition.
If i plug a = 3. I get x = 5. But, the range for x^2-3x-10<0 is x less than 5. Not including 5

Helly123 said:
If i plug a = 3. I get x = 5.
Really?

You shouldn't.

| x − 2 | < 3

gives

−3 < x − 2 < 3 .

Right?

SammyS said:
Really?

You shouldn't.

| x − 2 | < 3

gives

−3 < x − 2 < 3 .

Right?
Hm yes. Then, i am confused how to get 0 < a<= 3 itself..
How i get it is when x=-2, a must be 4
When x = 5, a must be 3
When x = 2, a must be 0
The range for a must be between 0 and 4

Helly123 said:
Hm yes. Then, i am confused how to get 0 < a<= 3 itself..
How i get it is when x=-2, a must be 4
When x = 5, a must be 3
When x = 2, a must be 0
The range for a must be between 0 and 4
You seem to be working this backwards.

If a = 4, then | x − 2 | < a gives −4 < x − 2 < 4, which includes values of x which do not satisfy the quadratic inequality.

Also, for a ≤ 0: There is no value of x for which | x − 2 | ≤ 0 . So we must have a > 0 .

Helly123 said:
Hm yes. Then, I am confused how to get 0 < a<= 3 itself.
@nuuskur outlined some of this in post #8.

If a ≤ 3 and | x − 2 | < a , then | x − 2 | < 3 .
...

You need to do some thoughtful work on these issues. We can't do it all for you.

Helly123
SammyS said:
You seem to be working this backwards.

If a = 4, then | x − 2 | < a gives −4 < x − 2 < 4, which includes values of x which do not satisfy the quadratic inequality.

Also, for a ≤ 0: There is no value of x for which | x − 2 | ≤ 0 . So we must have a > 0 .@nuuskur outlined some of this in post #8.

If a ≤ 3 and | x − 2 | < a , then | x − 2 | < 3 .
...

You need to do some thoughtful work on these issues. We can't do it all for you.
Ok i will try. Thank you

Is the necessary condition a>=4 or 0<a<=4 ?
Both is possible, i think

I must apologise for #8, it contains some false statements at the end.
A necessary condition is something that simply has to be true given some assumptions. In this case, we assume ##(x+2)(x-5)<0 ##
It does not imply that ##-1<x<5 ##, for instance, but it does imply ##-2<x<6 ##.

A quick way to check this, as I mentioned, is making use of the logical equivalence of two statements
$$P\implies Q\iff \neg Q\implies \neg P$$
In this example, as ## P## we consider ##(x+2)(x-5)<0 ## and as ##Q ## we consider ##\lvert x-2 \rvert <a ##.
If we want ##Q ## to be a necessary condition, we can also pick ## a## such that ##\neg Q = \lvert x-2\rvert \geq a ## is a sufficient condition for ##\neg P = (x+2)(x-5)\geq 0 ##.

It suffices to pick ##x\leq -2\lor x\geq 6 ##, then it's guaranteed that ##(x+2)(x-5)\geq 0 ##. Reverse the order now, if ##(x+2)(x-5)<0 ##, then it is necessarely true that ##-2<x<6 ## (i.e you are correct about ## a\geq 4##)

However, ##0<a<4 ## will not yield a necessary condition. If you pick, say, ## a=1##, then the implication
$$(x+2)(x-5)<0 \implies -1<x<3$$
is false. We can pick ##x = -1.5##, it still stands that ##(x+2)(x-5)<0 ##, but at the same time ##x\leq -1##.

I think I got confused by an earlier post, but I am certain now that this is correct.

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SammyS and Helly123
nuuskur said:
I must apologise for #8, it contains some false statements at the end.
A necessary condition is something that simply has to be true given some assumptions. In this case, we assume ##(x+2)(x-5)<0 ##
It does not imply that ##-1<x<5 ##, for instance, but it does imply ##-2<x<6 ##.

A quick way to check this, as I mentioned, is making use of the logical equivalence of two statements
$$P\implies Q\iff \neg Q\implies \neg P$$
In this example, as ## P## we consider ##(x+2)(x-5)<0 ## and as ##Q ## we consider ##\lvert x-2 \rvert <a ##.
If we want ##Q ## to be a necessary condition, we can also pick ## a## such that ##\neg Q = \lvert x-2\rvert \geq a ## is a sufficient condition for ##\neg P = (x+2)(x-5)\geq 0 ##.

It suffices to pick ##x\leq -2\lor x\geq 6 ##, then it's guaranteed that ##(x+2)(x-5)\geq 0 ##. Reverse the order now, if ##(x+2)(x-5)<0 ##, then it is necessarely true that ##-2<x<6 ## (i.e you are correct about ## a\geq 4##)

However, ##0<a<4 ## will not yield a necessary condition. If you pick, say, ## a=1##, then the implication
$$(x+2)(x-5)<0 \implies -1<x<3$$
is false. We can pick ##x = -1.5##, it still stands that ##(x+2)(x-5)<0 ##, but at the same time ##x\leq -1##.

I think I got confused by an earlier post, but I am certain now that this is correct.
nuuskur said:
I must apologise for #8, it contains some false statements at the end.
A necessary condition is something that simply has to be true given some assumptions. In this case, we assume ##(x+2)(x-5)<0 ##
It does not imply that ##-1<x<5 ##, for instance, but it does imply ##-2<x<6 ##.

A quick way to check this, as I mentioned, is making use of the logical equivalence of two statements
$$P\implies Q\iff \neg Q\implies \neg P$$
In this example, as ## P## we consider ##(x+2)(x-5)<0 ## and as ##Q ## we consider ##\lvert x-2 \rvert <a ##.
If we want ##Q ## to be a necessary condition, we can also pick ## a## such that ##\neg Q = \lvert x-2\rvert \geq a ## is a sufficient condition for ##\neg P = (x+2)(x-5)\geq 0 ##.

It suffices to pick ##x\leq -2\lor x\geq 6 ##, then it's guaranteed that ##(x+2)(x-5)\geq 0 ##. Reverse the order now, if ##(x+2)(x-5)<0 ##, then it is necessarely true that ##-2<x<6 ## (i.e you are correct about ## a\geq 4##)

However, ##0<a<4 ## will not yield a necessary condition. If you pick, say, ## a=1##, then the implication
$$(x+2)(x-5)<0 \implies -1<x<3$$
is false. We can pick ##x = -1.5##, it still stands that ##(x+2)(x-5)<0 ##, but at the same time ##x\leq -1##.

I think I got confused by an earlier post, but I am certain now that this is correct.
Where the -2<x<6 comes from?

If you pick ## a=4## then
$$\lvert x-2 \rvert < 4 \iff -4 < x-2 < 4 \iff -2 <x<6$$

Helly123
nuuskur said:
f you pick, say, ## a=1##, then the implication
$$(x+2)(x-5)<0 \implies -1<x<3$$
is false. We can pick ##x = -1.5##, it still stands that ##(x+2)(x-5)<0 ##, but at the same time ##x\leq -1##.
Ok, the implication is still false and the same counter-example works, but I should go back to primary school, because elementary arithmetic is, at times, an impossible task. It should be for ## a=1##
$$(x+2)(x-5)<0 \implies 1<x<3$$
Now, any ##x\in (-2,1]\cup [3,5) ## works as a counter-example.

but I am certain now that this is correct

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nuuskur said:
Ok, the implication is still false and the same counter-example works, but I should go back to primary school, because elementary arithmetic is, at times, an impossible task. It should be for ## a=1##
$$(x+2)(x-5)<0 \implies 1<x<3$$
Now, any ##x\in (-2,1]\cup [3,5) ## works as a counter-example.

but I am certain now that this is correct
Ok thanks

SammyS said:
Yes.
For | x − 2 | < a to be a sufficient condition so that x2 − 3x − 10 < 0, the range for a is 0 < a ≤ 3.
You have not been very clear in showing that this is the case.

This does not give all of the values of x which satisfy the given quadratic inequality. Rather any x fulfilling | x − 2 | < a will satisfy the quadratic inequality.for the specified range of a values. So being in that interval is a sufficient condition for x to satisfy the quadratic inequality..
I'm still hung up on the original problem statement which says x2 − 3x − 10 < 10. Did we decide somewhere that this was in error?

tnich said:
I'm still hung up on the original problem statement which says x2 − 3x − 10 < 10. Did we decide somewhere that this was in error?
Yes. I made mistake. It supposed to be x^2 - 3x - 10 < 0

## 1. What is the difference between necessary and sufficient conditions?

Necessary conditions are conditions that must be present in order for a certain outcome to occur. Sufficient conditions, on the other hand, are conditions that, if present, will guarantee the occurrence of a certain outcome. In other words, necessary conditions are required but not enough, while sufficient conditions are enough but not always required.

## 2. Can a necessary condition also be a sufficient condition?

Yes, a necessary condition can also be a sufficient condition in some cases. This means that if a certain condition is necessary for a particular outcome to occur, it is also enough for that outcome to occur. However, this is not always the case and it is important to distinguish between necessary and sufficient conditions.

## 3. How do necessary and sufficient conditions relate to causality?

Necessary conditions are often seen as the cause of a certain outcome, while sufficient conditions are seen as the effect. However, it is important to note that just because a necessary condition is present, it does not necessarily mean that the outcome will occur. It is also possible for multiple necessary conditions to exist for a single outcome.

## 4. Can necessary and sufficient conditions change over time?

Yes, necessary and sufficient conditions can change over time. This is because conditions can change and evolve, and what may have been a necessary or sufficient condition in the past may not be the case in the present. It is important for scientists to continually reassess and update their understanding of necessary and sufficient conditions.

## 5. How do scientists determine necessary and sufficient conditions?

Scientists determine necessary and sufficient conditions through observation, experimentation, and analysis of data. By carefully studying a phenomenon and identifying patterns and relationships, scientists can determine which conditions are necessary and which are sufficient for a certain outcome to occur. This process may involve trial and error and may require further research and analysis to reach a definitive conclusion.

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