🎓 Doc Freemo 🇳🇱 @freemo@qoto.org

@zeccano Your temporary silence has been removed.

If you continue to tag people after they ask you to go away multiple times the silence will be reinstated.

Be respectful, honor people's right to disengage.

For those who arent on an instance with math rendering you can read the answer here: http://mathb.in/38503

In the language of math the key is in how we frame the question. For example the following will provide some insight as I will show in a momemt.

\(\exists n \in \mathbb{N}\) such that \(n^3 - 1\) is prime

Not everyone understands the above notation so let me rephrase it more simply. The above translates to "There exists a Natural Number, \(n\), such that \(n^3 - 1\) is prime." Remember a Natural Numer is any positive Integer. In this example assuming that the original assertion that "8 is the only perfect cube to follow a prime" then \(n = 2\) which means the prime number, \(7\), is \(2^3 - 1 = 7\), and \(8\) is just \(2^3 = 8\). Easy enough, but how can we prove that this is the only case...

So we really just need to figure out which values for n in the equation above will give us a prime number, then we have our answer. A prime number is any number which only has 1 and itself as its factors. In other words the only two natural numbers we could possibly multiply together to get 7 is 1 and 7. So we have to start by factoring out the above equation \(n^3 - 1\), if we do that we get:

\[(n-1) \cdot (n^2+n+1)\]

It should be immediately obvious that of these two factors the left-most one is the smaller number, so we know:

\[(n-1) < (n^2+n+1)\]

Since we only care about prime numbers which satisfy the equation we know the left hand term must be equal to 1 and the right hand term most be equal to the entire number. So we can likewise assert the following:

\[1 = n-1\]
\[n^3 - 1 = n^2+n+1\]

Now we can use either equation and solve for n. It doesnt matter which equation you solve they will both give you the same value for n. Solving for n we get:

\[n = 2\]

If we plug that into the original equation, as we said earlier, er get the answer of 7, therefore 7 is the only prime number followed by a perfect cube.

\[2^3 - 1 = 7\]

One of my favorite math problems that is easy to solve with just algebra:

Prove that 8 is the only perfect cube to follow a prime number.

If you don't know what a perfect cube is, that is simple, it is any integer raised to the power of 3. Since \(8 = 2^3\) it is a perfect cube, and it follows the number 7, which is prime. 8 is the only number that fits those conditions... prove it.

NOTE: I will give the answer as a reply. If anyone else wants to provide an answer please make sure you use a content warning.

#maths #math #mathematics #riddle #puzzle #riddles #puzzles

Business Idea: Sell shirts soaked in oil and then washed really good. Then when you get food on your shirt and wash it it doesnt leave oil stains because the whole shirt is one big oil stain.

If anyone does this idea I demand royalties!

Seems I just hit the 7,000 follower mark.

Thank you to all the new followers!

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Newton was an asshole, team Leibniz!

#Leibniz4Life

When I was very young and new to math I remember being very fascinated with repeating decimals, things like this

\[0.121212121212...\]

for the rest of this post I will use the bar notation and represent the above as this which is the same thing:

\[0.\overline{12}\]

Anyway what I found curious as a kid was that you can represent an infinitely repeating decimal as a fraction where it is just the repeating bit of decimals over the same number of digits of 9. So the fraction form of the above looks like this

\[\frac{12}{99}\]

This blew my mind as a kid, it seemed almost magical how denominators of multiples of 10 were for "regular" decimals and denominators of 9 were for repeating decimals. What is this voodoo!

Of course it didnt take me long to put two and two together and realize that the fraction \(\frac{9}{9}\) results in 1 and not \(0.\overline{9}\) as the above naive rules might expect.

Of course I went to the math teacher to try to get some help understanding this and having a math teacher try to explain to a kid in first or second grade that \(0.\overline{9}\) is actually the same as \(1\) wasn't easy for him I'm sure.

I remember his logical point was "well you cant have an infinitely small number, so if the 9's go on forever they are just approaching 1". As an adult looking back I realize he was thinking of it in calculus terms, which is all well and valid, but it didnt translate too well to a little kid. I remember coming back tot he teacher the next day, smug as hell, ready to prove him wrong and I came up with this equation:

\[0.\overline{9} = 1 - 0.\overline{0}1\]

He said "you cant do that" I replied "I just did!"

As an adult knowing limits and all that I understand what the teacher was saying, and of course he was right in a way. But then even older as I got deeper in math I learned of the concept of an infinitesimal, which is basically the very idea I was conveying as a kid just in a less refined way, without good wording. So now I look back and dont know if my teacher was outsmarted by child-me, or what.

Me: All life is precious and beautiful and deserves our love and protection.

Them: What about humans?

Me: They can suck my dick!

Why is it every time I hear johnny cash since "Hurt" particularly the last recording of it he did before his death, it always makes me cry. I'm not even a Cash fan, and his voice wasnt at his best, it wasnt anything remarkable in terms of vocal talent... but it gets me every time.