Proving Irrational Numbers: Even Natural Numbers & Prime Products

[limitkiller]

Prove that:
1-If n^2 (n is a natural number) is even then n is even too .
2-Product of infinit number of primes bigger than 2 is not even.


Please do not "google it for me" .

 

[tiny-tim]

hi limitkiller!

show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[limitkiller]

hi limitkiller!

show us what you've tried, and where you're stuck, and then we'll know how to help!

I wanted too prove 1 (If n^2 (n is a natural number) is even then n is even too).

If n^2 is even then it is divisible by 2.

Then it has 2 as its divisors because it is an integer that divides n without leaving a reminder.

Then I said :Who says?, 2 (Product of infinit number of primes bigger than 2 is not even) might be incorrect .

 

[tiny-tim]

hi limitkiller!

I wanted too prove 1 (If n^2 (n is a natural number) is even then n is even too).

If n^2 is even then it is divisible by 2.

Then it has 2 as its divisors because it is an integer that divides n without leaving a reminder.

might be easier to prove the opposite …

start by assuming n is odd ​

Then I said :Who says?, 2 (Product of infinit number of primes bigger than 2 is not even) might be incorrect .

what can you say about a prime bigger than 2?

 

[limitkiller]

hi limitkiller!


might be easier to prove the opposite …

start by assuming n is odd ​

what can you say about a prime bigger than 2?

ummm I think I forgot to write one of the lines.Originally it should be like this:
I wanted too prove 1 (If n^2 (n is a natural number) is even then n is even too).

If n^2 is even then it is divisible by 2.

Then it has 2 as its divisors because it is an integer that divides n without leaving a reminder.

So n has 2 as its divisors too .

Then I said :Who says?, 2 (Product of infinit number of primes bigger than 2 is not even) might be incorrect .





About a prime bigger than 2 we can say its not divisible by 2, and product of a finit number of them is not divisible by 2...
But I couldn't find anything about product of infinit number of them.

 

[tiny-tim]

hi limitkiller!

Then it has 2 as its divisors because it is an integer that divides n without leaving a reminder.

So n has 2 as its divisors too .

yes, but you have to give a reason for that

(i still think my odd way is easier )

But I couldn't find anything about product of infinit number of them.

ah i see … i had assumed that was a misprint …

there's no such thing as a product of an infinite number of numbers , i think it must mean "a product of any size, however large"

 

[HallsofIvy]

I wanted too prove 1 (If n^2 (n is a natural number) is even then n is even too).

If n^2 is even then it is divisible by 2.

Then it has 2 as its divisors because it is an integer that divides n without leaving a reminder.

This proves that if a number is even, then its square is even. What you want to prove is the "converse": if the square of a number is even, then the number is even.

If n is odd then it is of the form 2k+ 1 for some integer k. What is the square of that?

Then I said :Who says?, 2 (Product of infinit number of primes bigger than 2 is not even) might be incorrect .

There is no such thing as a "product of an infinite number of primes". You can prove that the product of any number of primes, larger than 2, is odd.

Why was this thread titled "irrational numbers"?

 

[limitkiller]

there's no such thing as a product of an infinite number of numbers

Why?

Why was this thread titled "irrational numbers"?

Because (1) was used in the proof of "\sqrt{2} is not rational"

 

[HallsofIvy]

there's no such thing as a product of an infinite number of numbers

Because of the way multiplication is defined!
We define the product of two numbers, then define the product of three numbers as "xyz= (xy)z"- i.e. as the product of the two numbers xy and z (and use the associative property of multiplication to show that this is the same as x(yz)).

We can then inductively define the product of n+ 1 numbers as (x_1x_2x_3\cdot\cdot\cdot x_n)x_{n+1}. But that only gives us the product of N numbers for N a positive integer.

 

[limitkiller]

Then \prod_{k=1}^\infty k is meaningless.But it is not,is it?

 

[micromass]

Then \prod_{k=1}^\infty k is meaningless.But it is not,is it?

If you would like to give it a meaning, then the product would still diverge. So you can't really talk about divisibility...

 

[limitkiller]

then the product would still diverge. So you can't really talk about divisibility...

I don't get you . why does it mean that we can not talk about divisibility?

 

[micromass]

I don't get you . why does it mean that we can not talk about divisibility?

Because the product

\prod_{k=1}^{+\infty}{k}

diverges. It doesn't equal an integer, and you need integers to talk about divisibility.

 

[limitkiller]

Because the product

\prod_{k=1}^{+\infty}{k}

diverges. It doesn't equal an integer, and you need integers to talk about divisibility.

isn't it an integer itself?
for example 2^+\infty is divisible by 2.

 

[micromass]

isn't it an integer itself?
for example 2^+\infty is divisible by 2.

No: infinity is NOT an integer or a real number. You cannot calculate with infinity like you calculate with numbers.

At the very best, you might say that infinity is an extended real number. But that still doesn't allow you to talk about divisibility...

We have a good FAQ on the topic: https://www.physicsforums.com/showthread.php?t=507003

 

[limitkiller]

No: infinity is NOT an integer or a real number. You cannot calculate with infinity like you calculate with numbers.

At the very best, you might say that infinity is an extended real number. But that still doesn't allow you to talk about divisibility...

We have a good FAQ on the topic: https://www.physicsforums.com/showthread.php?t=507003

Thank you.
But I don't think I used infinity illegaly...
"Infinity is not a real number" does not mean that \sum_{k=1}^\infty \frac{1}{k} is not a real number or even \prod_{k=1}^\infty k is not a real number.

 

[jobsism]

But wouldn't it be possible to talk of certain properties of an infinite series?

I mean, for example, 1^(anything)=1. So, 1^(infinity) must also be 1, because regardless of the number of times we multiply, the answer will always be 1,right?

 

[micromass]

Thank you.
But I don't think I used infinity illegaly...
"Infinity is not a real number" does not mean that \sum_{k=1}^\infty \frac{1}{k} is not a real number or even \prod_{k=1}^\infty k is not a real number.

Both of those or not real numbers, whether you like it or not.

 

[micromass]

But wouldn't it be possible to talk of certain properties of an infinite series?

I mean, for example, 1^(anything)=1. So, 1^(infinity) must also be 1, because regardless of the number of times we multiply, the answer will always be 1,right?

No, 1^{+\infty} is an undetermined form (even in the extended real numbers). The reason is the sequence

\left(1+\frac{1}{n}\right)^n

Naively, you could say that 1+\frac{1}{n} converges to 1 and that n converges to infinity, and thus

\left(1+\frac{1}{n}\right)^n\rightarrow 1^{+\infty}=1

But this is NOT true! Check it for yourself: take a calculator and compute the first several terms in the sequence, it will not converge to 1!

The limit for the sequence is e and is defined as 2.718...

 

[limitkiller]

Both of those or not real numbers, whether you like it or not.

then why is \sum_{k=1}^\infty \frac{1}{k!} a real number?








I got it...

 

[micromass]

then why is \sum_{k=1}^\infty \frac{1}{k!} a real number?

Because that series converges.

 

[limitkiller]

Because that series converges.

Thanks

 

[Jocko Homo]

Thank you.
But I don't think I used infinity illegaly...
"Infinity is not a real number" does not mean that \sum_{k=1}^\infty \frac{1}{k} is not a real number or even \prod_{k=1}^\infty k is not a real number.

This is a surprisingly naive notion of infinity. It surprises me because I don't understand how you can know the terminology so well and even represent it competently with LaTeX but not understand what it is you yourself are saying...

In your summation example, what you are saying is shorthand for this:
\lim_{x\to\infty}\sum_{k=1}^x \frac{1}{k}
As you can see, this sequence grows without bound. It has no limit. That is to say, the limit doesn't exist. Real numbers exist. Therefore, the limit of this sequence can't be a real number. QED

You can peruse this thread about a mathematical paradox to get a better understanding of how limits differ from their... constructions, for lack of a better term...

Just because each successive approximation is a real number doesn't mean the limit must also be a real number...

 

[disregardthat]

If you extend the natural numbers by infinity, say that any unbounded sequence converges to infinity and that infinity * n = infinity for all natural numbers n, then you can extend the definition of divisibility for infinity. Clearly infinity is divisible by all numbers in this case.

 

[tiny-tim]

If you extend the natural numbers by infinity, say that any unbounded sequence converges to infinity and that infinity * n = infinity for all natural numbers n, then you can extend the definition of divisibility for infinity. Clearly infinity is divisible by all numbers in this case.

mmm … so 2^∞ is divisible by 3 ?