Proving the Existence of a Number with n Divisors

[Mathematicsresear]

Homework Statement

Prove that for every n greater than or equal to two, there exists a number with exactly n divisors.

Homework Equations

I used induction:

The Attempt at a Solution

Base case: assume n=2 which implies there exists a number with 2 divisors, that is the case for all prime numbers.
Now, let me assume that the statement is true for n=k, now I should show that it is true for n=k+1
which means that for every k+1, there exists an integers with exactly k+1 divisors,
k+1=(q_1...q_kq_k+1)
Now since k+1=(kq_k+1), therefore k divides k+1, and so does k+1 divides k+1.

Is this correct?

 

[fresh_42]

First of all, I assume you mean $n$ to be a natural number or an integer. This makes already a difference, as in one case $-1$ is a divisor, too, not to mentions rationals. Next, please define divisor, as we normally don't call units a divisor, which would exclude $1$ and a prime would be left as a number with only one divisor.

If all that is clear, why do you use induction? And your $k+1$ could have more than one divisor, which would lead to more than $n=k+1$ divisors which is not exactly $n$. Can you think of another way than induction to proceed? What special property has the number to have, which you add from one induction step to the next? If you figured that out, you don't need induction anymore.

 

[Mathematicsresear]

First of all, I assume you mean $n$ to be a natural number or an integer. This makes already a difference, as in one case $-1$ is a divisor, too, not to mentions rationals. Next, please define divisor, as we normally don't call units a divisor, which would exclude $1$ and a prime would be left as a number with only one divisor.

If all that is clear, why do you use induction? And your $k+1$ could have more than one divisor, which would lead to more than $n=k+1$ divisors which is not exactly $n$. Can you think of another way than induction to proceed? What special property has the number to have, which you add from one induction step to the next? If you figured that out, you don't need induction anymore.

An example of a divisors would be the following: if a is a divisor of b then aIb which means 0=amodb. Furthermore, the question states that the number has n divisors, so I assume it would mean that it is an integer. I am not sure on how to proceed after that.

 

[Dick]

Homework Statement

Prove that for every n greater than or equal to two, there exists a number with exactly n divisors.

Homework Equations

I used induction:

The Attempt at a Solution

Base case: assume n=2 which implies there exists a number with 2 divisors, that is the case for all prime numbers.
Now, let me assume that the statement is true for n=k, now I should show that it is true for n=k+1
which means that for every k+1, there exists an integers with exactly k+1 divisors,
k+1=(q_1...q_kq_k+1)
Now since k+1=(kq_k+1), therefore k divides k+1, and so does k+1 divides k+1.

Is this correct?

You seem to be confusing the number, the number of divisors and the number of prime factors. None of those things is necessarily equal. The divisors of 12 are 1, 2, 3, 4, 6 and 12. That's 6 divisors and the product of all of them isn't 12. Try and keep things clear. For a start, can you name a number with exactly 3 divisors?

 

[Mathematicsresear]

You seem to be confusing the number, the number of divisors and the number of prime factors. None of those things is necessarily equal. The divisors of 12 are 1, 2, 3, 4, 6 and 12. That's 6 divisors and the product of all of them isn't 12. Try and keep things clear. For a start, can you name a number with exactly 3 divisors?

4 has three divisors, 1,2 and 4.

 

[Dick]

4 has three divisors, 1,2 and 4.

Ok, now give me a number that has all of those divisors and exactly one more.

 

[Mathematicsresear]

Ok, now give me a number that has all of those divisors and exactly one more.

8 has divisors 1,2,4 and 8.

 

[Dick]

8 has divisors 1,2,4 and 8.

Does that suggest a way to do an induction step?

 

[Mathematicsresear]

Does that suggest a way to do an induction step?

I assume not, could you please explain?

 

[Dick]

I assume not, could you please explain?

How would you give me a number with 5 divisors based on what you've seen so far? Isn't that the same as the process of going from k to k+1? Which is what you want for induction.

 

[Mathematicsresear]

How would you give me a number with 5 divisors based on what you've seen so far? Isn't that the same as the process of going from k to k+1? Which is what you want for induction.

I don't think I am able to do such a thing.

 

[Dick]

I don't think I am able to do such a thing.

Here I thought we were doing so well. You can't continue the pattern to find a number with 5 divisors??

 

[Mathematicsresear]

Here I thought we were doing so well. You can't continue the pattern to find a number with 5 divisors??

I misunderstood, I meant I don't think I am able to use induction to do such a thing. 32 would be an example of a number with 5 divisors.

 

[SammyS]

... 32 would be an example of a number with 5 divisors.

No.
Count again.

 

[Mathematicsresear]

No.
Count again.

16 would be an example

 

[Ray Vickson]

Homework Statement

Prove that for every n greater than or equal to two, there exists a number with exactly n divisors.

Homework Equations

I used induction:

The Attempt at a Solution

Base case: assume n=2 which implies there exists a number with 2 divisors, that is the case for all prime numbers.
Now, let me assume that the statement is true for n=k, now I should show that it is true for n=k+1
which means that for every k+1, there exists an integers with exactly k+1 divisors,
k+1=(q_1...q_kq_k+1)
Now since k+1=(kq_k+1), therefore k divides k+1, and so does k+1 divides k+1.

Is this correct?

Questions:
(1) Do you exclude 1 as a divisor----almost everybody does in these situations.
(2) Do you mean $n$ different divisors? If not, you could take $N = 2^n$ as a number with exactly $n$ divisor, but this was probably not what was intended by the person posing the problem!

 

[Dick]

Questions:
(1) Do you exclude 1 as a divisor----almost everybody does in these situations.
(2) Do you mean $n$ different divisors? If not, you could take $N = 2^n$ as a number with exactly $n$ divisor, but this was probably not what was intended by the person posing the problem!

I really don't understand what your point is about 'different' divisors. Are you saying $2^n$ doesn't have $n$ (or $n+1$ depending on whether you count 1) different divisors?

 

[SammyS]

Questions:
(1) Do you exclude 1 as a divisor----almost everybody does in these situations.

I can't say whether it's "almost everybody" or it's "almost nobody", but following what @Dick points out, and I agree with him, in this case it makes sense to consider that 1 is a divisor.

(I see DIck beat me to it.)

 

[PeroK]

16 would be an example

From this thread, I would suggest you haven't completely grasped the concept of induction. For this problem, in fact, you do not need induction. You could try a direct proof:

Let $n \ge 2$, now all you have to do is find a number with exactly $n$ divisors.