Can we justify selecting an integer in between two reals in Rudin 1.20 Theorem?

[Unassuming]

This proof seems amazing. I finally understood it after one hour. (I am slow and not mature)

I have a question though.

In 1.20 (b),
The line that says, "Hence there is an integer m (with -m₂$$\leq$$m$$\leq$$m₁) such that...

How does he justify selecting an m in between the two? I feel uncomfortable with this since it is similar to what is being proved, except for the fact that m is an integer.

Is there a Theorem that says we can pick an integer in between two reals?

 

[NoMoreExams]

I don't have Rudin but of course you can't find an integer between 2 Reals, pick x = 1.2 and y = 1.21, can you find an integer in between?

 

[Count Iblis]

Is m1 + m2 > 1 ?

 

[Integral]

It would help those that do not have a copy of Rudin if you were to actually state what the theorem is.

Here is an exersise for you, find a pair of integers, m₁>0 and m₂>0 such that there is NO integer between m₁ and - m₂ .

 

[Unassuming]

Pg 9

If x in R, y in R, and x<y, then there exists a p in Q such that x<p<y.

Proof:
Since x<y, we have y-x>0, and the archimedean property furnishes a positive integer n such that
n(y-x)>1.

Apply the archimedean property again, to obtain positive integers m₁ and m₂ such that m₁>nx, m₂>-nx. Then

-m₂<nx<m₁.

Hence there is an integer m ( with -m₂<= m <= m₁) such that

m-1 <= nx < m

If we combine these inequalities, we obtain

nx<m<= 1 + nx < ny.

Since n>0, it follows that

x < m/n < y. This proves (b) with p = m/n.

 

[Unassuming]

It would help those that do not have a copy of Rudin if you were to actually state what the theorem is.

Here is an exersise for you, find a pair of integers, m₁>0 and m₂>0 such that there is NO integer between m₁ and - m₂ .

Sorry to those who wanted to see the entire proof before but I did not know if it was "legal". Anyway, I put it up.

As for the "exersise" assigned to me...

 

[morphism]

I don't really see what the problem is. m₁ and m₂ are positive integers, so there are finitely many integers between -m₂ and m₁, namely -m₂, -m₂+1, ..., m₁. Rudin then picks the first m that comes right after nx, and we know that this will lie between -m₂ and m₁ (inclusive), because -m₂ < nx < m₁.

 

[JinM]

Then

-m₂<nx<m₁.

Hence there is an integer m ( with -m₂<= m <= m₁) such that

m-1 <= nx < m

I was looking at that proof today, and well, I don't exactly understand where the last inequality, m - 1 <= nx < m, came from. In particular, why get m_1 and m_2 at all if all we're looking for is m? Also, how do you guarantee that m - 1 <= nx? Maybe this proof is too slick for me, but I appreciate any pointers.

 

[Unassuming]

I was looking at that proof today, and well, I don't exactly understand where the last inequality, m - 1 <= nx < m, came from. In particular, why get m_1 and m_2 at all if all we're looking for is m? Also, how do you guarantee that m - 1 <= nx? Maybe this proof is too slick for me, but I appreciate any pointers.

I am no expert but I have tried very hard to understand this one. Let me know if this helps.

He picked m with 2 conditions: (1) it must be in between m1 and m2
(2) it must be within "1 unit" distance from nx, such that m-1 is less than nx.

He might have needed the m1 and m2 because we cannot just pick any number. We therefore used archidean property to find m1 and m2, and then again to find a m in between them but a little to the "right" of nx.

Also, if m is within "1 unit" distance from nx, then "nx+1" shall overpass m on the right.

so, m-1 < nx < m < nx+1.

 

[JinM]

Thanks Unassuming. I learned from the IRC channel that the omission is simply the use of the well-ordering principle: that is, any nonempty subset of N must have a minimum. Take the set {n \in N, nx < n}, it is nonempty since we have m_2 > nx. By the well ordering principle, it also must have a minimum, call it m. Then m - 1 is clearly less than nx, since m is the minimum element for which the condition in the above set holds, and you get your inequality.