You are Here: Home >< Maths

Clever way of showing which of these two fractions is greater? watch

Announcements
1. Anyone got a nifty way to show which of the following is larger?

1) 1/(1+x)

2) (1/x) - 1
2. (Original post by Economister)
Anyone got a nifty way to show which of the following is larger?

1) 1/(1+x)

2) (1/x) - 1
For what x?
3. Suppose 1) = 2)

1/(1+x) = (1/x) - 1

1 = (x+1)/x - 1 - x
1 = 1 + 1/x - 1 - (x^2)/x
x = -x^2 + 1
x^2 + x - 1 = 0

the roots of that are

-1.61803

and

0.61803

If x > 0.61803 , 1) > 2)

If -1.61803 < x < 0 then 1) > 2)
If 0 < x < 0.61803 then 2) > 1)

If x < -1.61803 then 1) > 2)

That's the best that can be done. Neither is DEFINITELY larger than the other. But 1) is much more likely to be bigger as 2) is only bigger in a very small range.

Also they are not comparable if x = 0 or x = -1 as one will be infinite.

EDIT: I think I have corrected my inequalities now
4. (Original post by hassi94)
Suppose 1) = 2)

1/(1+x) = (1/x) - 1

1 = (x+1)/x - 1 - x
1 = 1 + 1/x - 1 - (x^2)/x
x = -x^2 + 1
x^2 + x - 1 = 0

the roots of that are

-1.61803

and

0.61803

If x > 0.61803 , 1) > 2)

If -1.61803 < x < 0.61803 then 2) > 1)

If x < -1.61803 then 1) > 2)

That's the best that can be done. Neither is DEFINITELY larger than the other. But 1) is much more likely to be bigger as 2) is only bigger in a very small range.

Also they are not comparable if x = 0 or x = -1 as one will be infinite.
A rather dangerous statement me-thinks ...... I could equally argue that either statement is true for an infinite "number" of values ....
5. (Original post by H.C. Chinaski)
A rather dangerous statement me-thinks ...... I could equally argue that either statement is true for an infinite "number" of values ....
Yeah you're right, but he's doing economics not maths so it would be applied more to real life and that's why I derived that statement.

But yeah I agree probably not the best statement to make
6. (Original post by hassi94)
If -1.61803 < x < 0.61803 then 2) > 1)
Take x=-0.1:

1/(1+(-0.1)) = 1.1111....

1/(-0.1) - 1 = -11

2) < 1)

The mistake you made was you jumped from talking about equalities to inequalities without properly considering the signs of everything you multiply each side by, causing the inequality to flip round.

The approach I would take would be to consider 1/x - 1 - 1/(1+x) and see when it's positive, negative or undefined.
Spoiler:
Show
It helps to add fractions and factorise
7. 1) 1/(1+x)

2) (1/x) - 1
8. (Original post by ttoby)
Take x=-0.1:

1/(1+(-0.1)) = 1.1111....

1/(-0.1) - 1 = -11

2) < 1)

The mistake you made was you jumped from talking about equalities to inequalities without properly considering the signs of everything you multiply each side by, causing the inequality to flip round.

The approach I would take would be to consider 1/x - 1 - 1/(1+x) and see when it's positive, negative or undefined.
Spoiler:
Show
It helps to add fractions and factorise
Yeah sorry my bad just realised how wrong that was haha. Don't know why I didn't consider the fact I was multiplying by an unknown quantity and how that could affect the inequality.

9. (Original post by hassi94)
Suppose 1) = 2)

1/(1+x) = (1/x) - 1

1 = (x+1)/x - 1 - x
1 = 1 + 1/x - 1 - (x^2)/x
x = -x^2 + 1
x^2 + x - 1 = 0

the roots of that are

-1.61803

and

0.61803

If x > 0.61803 , 1) > 2)

If -1.61803 < x < 0.61803 then 2) > 1)

If x < -1.61803 then 1) > 2)

That's the best that can be done. Neither is DEFINITELY larger than the other. But 1) is much more likely to be bigger as 2) is only bigger in a very small range.

Also they are not comparable if x = 0 or x = -1 as one will be infinite.
I think you have the range of values incorrect possibly because you haven't use inequalities and somewhere you may have got a sign wrong - I haven't checked fully though. For this, I would use the sign test (but before you do this you will have to collect fractions and make a table of the critical points).

Edit: LOL posted at the same time - oh well
10. (Original post by ttoby)
Take x=-0.1:

1/(1+(-0.1)) = 1.1111....

1/(-0.1) - 1 = -11

2) < 1)

The mistake you made was you jumped from talking about equalities to inequalities without properly considering the signs of everything you multiply each side by, causing the inequality to flip round.

The approach I would take would be to consider 1/x - 1 - 1/(1+x) and see when it's positive, negative or undefined.
Spoiler:
Show
It helps to add fractions and factorise
I've corrected it now I believe.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 9, 2011
Today on TSR

University open days

1. Norwich University of the Arts
Fri, 28 Sep '18
2. Edge Hill University
Faculty of Health and Social Care Undergraduate
Sat, 29 Sep '18
3. Birmingham City University
Sat, 29 Sep '18
Poll
Useful resources

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams

Chat with other maths applicants