MHB Is there an other way to show the proposition ?

[evinda]

Hello! (Wave)

I am looking at the proof that $\prod_{p \leq 2n}p>2^n, \forall n \geq 2$, where the product extends over all primes $p \leq 2n$.

For $n<100$ the above proposition is shown as follows.

View attachment 7678

I was wondering if there is also an other way to prove the proposition for $n<100$.

Also is it a formal proof to just check that the proposition is true for some values $n<100$, as it is done at the picture above? (Thinking)

 

[I like Serena]

I was wondering if there is also an other way to prove the proposition for $n<100$.

Hey evinda! (Smile)

I'm afraid that I wouldn't know.
If we could, perhaps we could prove it for all $n$ in general! (Happy)

Also is it a formal proof to just check that the proposition is true for some values $n<100$, as it is done at the picture above? (Thinking)

For a formal proof up to $n<100$, we need to verify it for all those values of $n$.
So let's take a look if we missed one.
We seem to be missing $n=4$, so let's see...
For $n=3$ we have that $\Pi p > 2^4$, so I think that for $n=4$ we will also have that $\Pi p > 2^4$ won't we?
So then we didn't miss $n=4$ after all. (Thinking)

 

[evinda]

For a formal proof up to $n<100$, we need to verify it for all those values of $n$.
So let's take a look if we missed one.
We seem to be missing $n=4$, so let's see...
For $n=3$ we have that $\Pi p > 2^4$, so I think that for $n=4$ we will also have that $\Pi p > 2^4$ won't we?
So then we didn't miss $n=4$ after all. (Thinking)

I see...
Then, is it right as follows?

For $n=8$ we have that $\Pi p > 2^{14}$, so for $n=9, \dots, 14$ we will also have that $\Pi p > 2^{14}$ .

For $n=15$ we have that $\Pi p > 2^{30}$, so for $n=16, \dots, 30$ we will also have that $\Pi p > 2^{30}$ .

For $n=30$ we have that $\Pi p > 2^{65}$, so for $n=31, \dots, 65$ we will also have that $\Pi p > 2^{65}$ .

For $n=65$ we have that $\Pi p > 2^{149}$, so for $n=66, \dots, 99$ we will also have that $\Pi p > 2^{149}$ .For example for $n=99$ we want to have $\Pi p>2^{90}$ which is true since it holds that $\Pi p > 2^{149}$.

Am I right? (Thinking)

 

[I like Serena]

I see...
Then, is it right as follows?

For $n=8$ we have that $\Pi p > 2^{14}$, so for $n=9, \dots, 14$ we will also have that $\Pi p > 2^{14}$ .

For $n=15$ we have that $\Pi p > 2^{30}$, so for $n=16, \dots, 30$ we will also have that $\Pi p > 2^{30}$ .

For $n=30$ we have that $\Pi p > 2^{65}$, so for $n=31, \dots, 65$ we will also have that $\Pi p > 2^{65}$ .

For $n=65$ we have that $\Pi p > 2^{149}$, so for $n=66, \dots, 99$ we will also have that $\Pi p > 2^{149}$ .For example for $n=99$ we want to have $\Pi p>2^{90}$ which is true since it holds that $\Pi p > 2^{149}$.

Am I right? (Thinking)

Yep. (Nod)

It seems that whoever wrote the proof, continued each time where the condition might be violated, but wasn't violated after all. (Nerd)

 

[evinda]

Yep. (Nod)

It seems that whoever wrote the proof, continued each time where the condition might be violated, but wasn't violated after all. (Nerd)

Yes, for example after $n=15$ we could check the value $n=31$ instead of $n=30$, right? (Thinking)

Also, do you maybe have an idea what is meant with the following?

For establishing the table we have used the fact that between $30$ and $60$ there are $7$ prime numbers ( and that $31 \cdot 37>2^{10}$), between $60$ and $130$ there are $14$.
How do we use these facts? (Thinking)

 

[I like Serena]

Yes, for example after $n=15$ we could check the value $n=31$ instead of $n=30$, right?

Yep. (Nod)

Also, do you maybe have an idea what is meant with the following?

For establishing the table we have used the fact that between $30$ and $60$ there are $7$ prime numbers ( and that $31 \cdot 37>2^{10}$), between $60$ and $130$ there are $14$.
How do we use these facts?

Don't we have
$$\displaystyle\prod_{p<2\cdot 15} p > 2^{30}$$
so that
$$\displaystyle\prod_{p<2\cdot 30} p \quad=\quad \prod_{p<2\cdot 15} p \cdot \prod_{2\cdot 15 \le p<2\cdot 30} p\quad>\quad 2^{30} \cdot \prod_{2\cdot 15 \le p<2\cdot 30} p \quad=\quad 2^{30} \cdot 31 \cdot ... \cdot 59$$
(Wondering)

 

[evinda]

Yep. (Nod)
Don't we have
$$\displaystyle\prod_{p<2\cdot 15} p > 2^{30}$$
so that
$$\displaystyle\prod_{p<2\cdot 30} p \quad=\quad \prod_{p<2\cdot 15} p \cdot \prod_{2\cdot 15 \le p<2\cdot 30} p\quad>\quad 2^{30} \cdot \prod_{2\cdot 15 \le p<2\cdot 30} p \quad=\quad 2^{30} \cdot 31 \cdot ... \cdot 59$$
(Wondering)

I see... but do we need this for establishing the table?
Doesn't $\displaystyle\prod_{p<2\cdot 15} p > 2^{30}$ directly imply that $\displaystyle\prod_{p<2\cdot 30} p \geq \displaystyle\prod_{p<2\cdot 15} p >2^{30}$ ? (Thinking)

 

[I like Serena]

I see... but do we need this for establishing the table?
Doesn't $\displaystyle\prod_{p<2\cdot 15} p > 2^{30}$ directly imply that $\displaystyle\prod_{p<2\cdot 30} p \geq \displaystyle\prod_{p<2\cdot 15} p >2^{30}$ ? (Thinking)

Indeed, for $n=30$ we do not need it.
But for $n=31$ up to $n=60$ we do, don't we? (Wondering)
And it actually works up to $n=65$, since the table lists $2^{65}$ for it.

 

[evinda]

But for $n=31$ up to $n=60$ we do, don't we? (Wondering)
And it actually works up to $n=65$, since the table lists $2^{65}$ for it.

Why? I haven't understood it... (Worried)I thought that it would be as follows.

$\prod_{p \leq 2 \cdot 30} p=2^{65}>2^{64}$ and so $\prod_{p \leq 2 \cdot \lambda} p>2^{64}, \forall \lambda \in \{ 31,\dots, 64\}$. Am I wrong? (Thinking)

 

[I like Serena]

Why? I haven't understood it... (Worried)I thought that it would be as follows.

$\prod_{p \leq 2 \cdot 30} p=2^{65}>2^{64}$ and so $\prod_{p \leq 2 \cdot \lambda} p>2^{64}, \forall \lambda \in \{ 31,\dots, 64\}$. Am I wrong? (Thinking)

Huh?
It seems to me that you did understand it!
It's only that instead of $\prod_{p \leq 2 \cdot 30} p=2^{65}$ we have:
$$\displaystyle\prod_{p\le 2\cdot 30} p \quad=\quad \prod_{p\le 2\cdot 15} p \cdot \prod_{2\cdot 15 < p \le 2\cdot 30} p\quad>\quad 2^{30} \cdot \prod_{2\cdot 15 < p\le 2\cdot 30} p \quad=\quad 2^{30} \cdot 31 \cdot ... \cdot 59 \quad>\quad 2^{65}$$
So:
$$\displaystyle\prod_{p\le 2\cdot 30} p > 2^{65}$$
Can you clarify? (Wondering)

 

[evinda]

Huh?
It seems to me that you did understand it!

So is that what I wrote in post #9 correct? (Thinking)

It's only that instead of $\prod_{p \leq 2 \cdot 30} p=2^{65}$ we have:
$$\displaystyle\prod_{p\le 2\cdot 30} p \quad=\quad \prod_{p\le 2\cdot 15} p \cdot \prod_{2\cdot 15 < p \le 2\cdot 30} p\quad>\quad 2^{30} \cdot \prod_{2\cdot 15 < p\le 2\cdot 30} p \quad=\quad 2^{30} \cdot 31 \cdot ... \cdot 59 \quad>\quad 2^{65}$$
So:
$$\displaystyle\prod_{p\le 2\cdot 30} p > 2^{65}$$

I see... (Nod)

 

[I like Serena]

So is that what I wrote in post #9 correct?

Yep. (Nod)