Can Natural Numbers m and n Satisfy x + 1/m < y - 1/n?

[mathstudent1]

Homework Statement

prove x+1/m < y-1/n

Relevant Equations

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Let x, y ∈ R be such that x < y. Prove that there exist natural numbers m and n such that x +1/m < y −1/n?

 

[FactChecker]

Because this is a homework-type of problem, you must show your work and then we can give hints and guidance. I will cheat and give you an initial hint. Consider y-x and see what that tells you about n and m.

 

[PeroK]

Can you make an attempt at this?

 

[mathstudent1]

Because this is a homework-type of problem, you must show your work and then we can give hints and guidance. I will cheat and give you an initial hint. Consider y-x and see what that tells you about n and m.

 

[mathstudent1]

i try but i do not know how to complete and i think my assumption is false

 

[mathstudent1]

Can you make an attempt at this?

i think my assumption is false i don ot know how i complete

 

[PeroK]

Please type yoru attempt into a post, with some explanation of what you are doing.

 

[docnet]

For your proof to work, you would need to begin with $\frac{1}{m}+\frac{1}{n}<0$, which is false for any $m,n>0$.

If you get stuck, try proving it differently. Can you choose $m$ and $n$ such that $\frac{1}{m}+\frac{1}{n}$ is less than $y-x$?

 

[mathstudent1]

by archimedean property right ? there is n,m which are integers s.t n>1/y and 1/m>-1/x then 1/n<y and 1/m<-x , add them we will get 1/m+1/n <y-x.

 

[PeroK]

by archimedean property right ? there is n,m which are integers s.t n>1/y and 1/m>-1/x then 1/n<y and 1/m<-x , add them we will get 1/m+1/n <y-x.

At some point, you'll need to learn to structure your arguments properly. That argument doesn't work as $m, n$ must be natural numbers. In particular, $\frac 1 m$ and $\frac 1 n$ cannot be negative.

 

[mathstudent1]

At some point, you'll need to learn to structure your arguments properly. That argument doesn't work as $m, n$ must be natural numbers. In particular, $\frac 1 m$ and $\frac 1 n$ cannot be negative.

you are right , then how i prove it ?

 

[PeroK]

you are right , then how i prove it ?

My general advice if you are not sure how to prove something is to justify that it is true. In this case, you could start with, say, $x = 0$ and $y = 1$ and find suitable $m, n$. Can you do that to start with?

Note that if you were more experienced, you might see a way to generalise that simple case to cover the case for any $x$ and $y$.

 

[mathstudent1]

yes i can take m=3 and n=2

 

[PeroK]

yes i can take m=3 and n=2

If $y - x \ge 1$, then those would do in general.

What about the case where $y - x < 1$?

 

[mathstudent1]

If $y - x \ge 1$, then those would do in general.

What about the case where $y - x < 1$?

they said in question x<y ,

If $y - x \ge 1$, then those would do in general.

What about the case where $y - x < 1$?

also i can found m and n but i do not know who i sketch the proof

 

[PeroK]

they said in question x<y ,

Yes, but if $x < y$, then $y - x > 0$. That splits into two cases:

Case 1: $y - x \ge 1$, where you have the solution $m = \frac 1 3, n = \frac 1 2$.

Case 2: $0 < y-x < 1$. Which looks harder, because $y-x$ is "small".

 

[FactChecker]

Rewrite x+1/m < y-1/n as 1/m+1/n < y-x by subtracting x from both sides and adding 1/n to both sides.
So the problem can be restated as proving that you can find Natural numbers n,m such that 1/m+1/n < y-x.
Now, how can you make 1/m+1/n small enough? (Remember that x < y, so 0 < y-x.)

 

[PeroK]

also i can found m and n but i do not know who i sketch the proof

Let me give you a bit more help. When $x =0$ and $y = 1$, here's what I think you probably did:

Note that $\frac 1 2$ is half way between $0$ and $1$. So, let's take $m, n \ge 2$. Although, it might have been better to take $m = n = 3$.

For general $x, y$ you could generalise this:

Note that $\frac {y- x}{2}$ is half the distance between $x$ and $y$. This is where you can apply the Archimedean property and take $\frac 1 m = \frac 1 n < \frac{y-x}{2}$.

 

[SammyS]

Let me give you a bit more help. When $x =0$ and $y = 1$, here's what I think you probably did:

Note that $\frac 1 2$ is half way between $0$ and $1$. So, let's take $m, n \le \frac 1 2$. Although, it might have been better to take $m = n = \frac 1 3$.
. . .

@PeroK ,
Don't you mean $\displaystyle \frac 1 m$ and $\displaystyle \frac 1 n$ rather than $m$ and $n$ in this post ?

 

[pasmith]

Rewrite x+1/m < y-1/n as 1/m+1/n < y-x by subtracting x from both sides and adding 1/n to both sides.
So the problem can be restated as proving that you can find Natural numbers n,m such that 1/m+1/n < y-x.
Now, how can you make 1/m+1/n small enough? (Remember that x < y, so 0 < y-x.)

You can take n = m, so that you need \frac1{n} &lt; \frac{y - x}{2}.

 

[PeroK]

@PeroK ,
Don't you mean $\displaystyle \frac 1 m$ and $\displaystyle \frac 1 n$ rather than $m$ and $n$ in this post ?

Yes. Fixed, thanks.

 

[FactChecker]

You can take n = m, so that you need \frac1{n} &lt; \frac{y - x}{2}.

That is a nice simplification that allows one to get a formula for n (=m).
$n > \frac {2}{y-x}$, and $m=n$.

 

[pasmith]

It is sufficient to require <br /> \frac1n + \frac1m \leq \frac 2{\min(n,m)} &lt; y - x where the first inequality is true for all positive n and m. This again leads to the problem of proving the existence of a positive N such that N &gt; \frac{2}{y - x}.