The infimum and supremum of a set containing rational numbers whose squares satisfies a certain property.

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Solution 1

I think that $[{}\cdot {}]$, as it appears in the first numerator in the definition of $w$, is the floor function (also called the greatest integer function). That makes $[(n+1)s]$ an integer (the greatest integer $m$ such that $m \leq [(n+1)s]$, by definition), so all numerators and denominators in the definition of $w$ are integers, which makes $w$ rational.

Solution 2

Another way to approach, e.g., the question of what the maximum is in this set: You are searching for rational numbers $p/q$ with both $p, q > 0$ for which $p^2 < 2q^2$. The last inequality is equivalent to $(p/q)^2 < 2$, which, in turn, is equivalent to $p/q < \sqrt{2}$.

So, you are trying to find the maximum rational $p/q$ that is less than $\sqrt{2}$; but clearly this does not exist, for you can take a rational sequence that is monotonically increasing and converges to $\sqrt{2} \not\in \mathbb{Q}$, which means that for any rational less than $\sqrt{2}$ you can go far out enough in this sequence to find an even greater rational that is still less than $\sqrt{2}$.

A concrete example of such a sequence can be found by looking at the decimal expansion of $\sqrt{2}$:

$$\sqrt{2} = 1.414213\ldots$$

Next, use rationals of the form "natural number" over "power of ten" corresponding to the above:

$$1/1, 14/10, 141/100, 1414/1000, 14142/10000, 141421/100000, 1414213/1000000, \ldots$$

Using this monotonically increasing sequence that converges to $\sqrt{2}$, we find that the maximum of the given set does not exist.

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Emptymind

Updated on August 01, 2022

• Emptymind over 1 year

Find the infimum,supremum,maximum and minimum of the following set or prove non existence: $$E = \{\frac{p}{q} \in \mathbb{Q}: p^{2} < 2q^{2}, p,q >0 \}.$$

This question is solved in the following:

But I did not understand why he said that $\omega$ (as he defined in the second picture) is a rational number, Could anyone explain this for me?

• erfink over 6 years
$[ x]$ is being used to denote the floor function, which returns the largest integer that is smaller than $x$.
• erfink over 6 years
Please see math.meta.stackexchange.com/questions/5020/… for help with typesetting; posting screenshots or pictures of math is generally considered poor form.
• Emptymind over 6 years
I am so sorry @erfink .
• erfink over 6 years
No worries, just for future reference =) math.meta.stackexchange.com/questions/13677/… for more discussion
• DanielWainfleet over 6 years
There are simpler shorter proofs. See Heron's (Hero's) Fomula for square roots, for example. And a little bit of Number Theory makes proving that $\sqrt 2$ is irrational rather easy.
• Emptymind over 6 years
it is the step(Dirichlet) function.
• Arthur over 6 years
@MathLover When I google "Dirichlet function", I get something entirely different, and the step function is just $0$ for negative input and $1$ for positive input, so I don't think that's what's used here either. What function do you mean, explicitly?
• Emptymind over 6 years
my mother tongue is not English,may be I said the name wrong sorry, but this function means to take the integer value that is nearer to your number, this what I understand.
• Arthur over 6 years
Ahh, ok. So it's not specifically rounding down like the floor function does, but rather just rounding to the closest integer. No matter, $[(n+1)s]$ is still an integer.
• Emptymind over 6 years
yeah Thank u :), but how can we find the nearer integer for the product of an integer $(n+1)$ and a number that we do not know if it is an integer or no which is s?
• fleablood over 6 years
Needs to be the floor so we know that s < w < s + 1/n if we round up we could have w > s + 1/(n+1) and we could have w >= s + 1/n. Bur if we round down we will have s < w <= s + 1/(n+1) < s + 1/n. But both ways w must be rational.
• Arthur over 6 years
We don't have to find a number in order to talk about it; that's why they write it using the rounding function. What you're asking is like asking how we can find $s$ when we don't know whether it's a rational or an irrational number. The answer is we don't; we know the number exists, we know some of the properties it has, and we attach a name to it. In the latter case, that name is $s$, and in the former case, that name is $[(n+1)s]$, which is a bit longer because it is not just a name, it contains a description of the aforementioned properties, but it is still as valid as a single letter.
• fleablood over 6 years
For every real number $x$ there is a unique integer, $k$ so that $k \le x < k+1$. That is true for all real numbers. We notate $k$ as $[x]$. so for $(n+1)s$ there is a unique integer $k$ so that $k \le (n+1)s < k + 1$. We denote that integer as $k = [(n+1)s]$.