Comparing 2 Fraction based variables

[zak100]

Homework Statement

c & d are positive integers &
1/c = 1 + 1/d
Now tell which is greater c or d?

Homework Equations

1/c = 1 + 1/d

The Attempt at a Solution

1/c = (d+1)/d

c = d/ (d+1)
c = 1 + d

Thus if c =10
therefore d = 9
But this is not a correct answer. Some body please guide me.

Zulfi.

 

[Nidum]

Hint : Which is smaller 1/c or 1/d ?

 

[rpthomps]

You could also just sub in values and make inferences from that

 

[Charles Link]

For c, d positive integers, 1/c is always less than 1 unless c=1. I think the statement of the problem is incorrect that c and d are positive integers. There is no solution. c=1 implies d=+infinity. If the problem statement had said positive reals, then in one case c<1 and d>1. In another case, c<1 and d<1 and c<d, so that c<d if c, d are positive reals.

 

[Ray Vickson]

Homework Statement

c & d are positive integers &
1/c = 1 + 1/d
Now tell which is greater c or d?

Homework Equations

1/c = 1 + 1/d

The Attempt at a Solution

1/c = (d+1)/d

c = d/ (d+1)
c = 1 + d

Thus if c =10
therefore d = 9
But this is not a correct answer. Some body please guide me.

Zulfi.

As pointed out in #4, the equation is impossible if c and d are positive integers; however, if they are just positive real numbers, then the equation says that 1/c is a bigger number than 1/d. What does that say about c and d themselves?

 

[zak100]

Hi,
Thanks for your help. Actually it says that " c & d are positive".

<Which is smaller 1/c or 1/d ?>
I think it depends on values. They can be equal.

<
In another case, c<1 and d<1 and c<d, so that c<d if c, d are positive reals.>
If c & d are positive reals this means that c & d can be integers also.
What's wrong with my solution. I have proved that c is greater. However the answer says that d is greater.

Zulfi.

 

[Charles Link]

Hi,
Thanks for your help. Actually it says that " c & d are positive".

<Which is smaller 1/c or 1/d ?>
I think it depends on values. They can be equal.

<
In another case, c<1 and d<1 and c<d, so that c<d if c, d are positive reals.>
If c & d are positive reals this means that c & d can be integers also.
What's wrong with my solution. I have proved that c is greater. However the answer says that d is greater.

Zulfi.

Your last line of post #1 c=1+d is incorrect. Meanwhile, since 1/c=1+1/d, this says the right side of the equation is greater than 1. That means c must be less than 1 (to make the left side greater than 1) and therefore c can not be a positive integer.

 

[Nidum]

I guessed that the use of integer was a typo - the question did not make sense otherwise .

Now think about this :

If x is greater than y then is 1/x greater or less than 1/y ? Put some numbers in if it helps - say 10 for x and 5 for y .

 

[zak100]

Hi,
Thanks for your response. I got my mistake. I am now able to understand book's solution.
According to the book,
c = d/ (d+1)
Since d is positive, so d+1> d. if d= 7 then 7/(7+1) (which is a fraction & less than d). So d > d/(d+1). But c=d/(d+1). So d>c.

Now your question:
"If x is greater than y then is 1/x greater or less than 1/y ? Put some numbers in if it helps - say 10 for x and 5 for y" .

i.e. x> y so 1/x < 1/y. Inequalities reverse. 10> 5 so 1/10 < 1/5.

Thanks for help.

Zulfi.

 

[zak100]

Hi,
I have a question with my original solution:
i.e
1/c = 1 + 1/d
Why can't we write:
c= 1 + d?
which is obtained by taking reciprocal of both sides.

Zulfi.

Reference https://www.physicsforums.com/threads/comparing-2-fraction-based-variables.875524/

 

[Mark44]

Hi,
I have a question with my original solution:
i.e
1/c = 1 + 1/d
Why can't we write:
c= 1 + d?

Because it is incorrect. See below.

which is obtained by taking reciprocal of both sides.

The reciprocal of ##1 + \frac 1 d## is NOT 1 + d.
Here is an example using numbers.
$$\frac 1 {\frac 1 3} = 1 + \frac 1 {\frac 1 2}$$
Does it follow that ##\frac 1 3 = 1 + \frac 1 2##?

Zulfi.

Reference https://www.physicsforums.com/threads/comparing-2-fraction-based-variables.875524/