@Xam -- Well, I frankly don't know. I'm not planning to pursue a math degree, so I'm probably not into math as others. However, I do like geometry (I'm currently trying to solve a puzzle given to me by a mathematician). But I really can't wait until I learn Taylor series.
Well, this is something basic but I can not figure out the formal argument. Let's say we have a matrix $ M $ such that there is exactly one $ 1 $ in each column and in each row. Then $ M $ is a square matrix.
Clearly, we are going to have a crazy time trying to diagonolize this. Anyways, we move on to the next 'level'
This level when put into the fast growing hierarchy produces functions that grow so fast, you can't do basic PA with it: $$\epsilon_0=\sup\{\omega,\omega^\omega, \omega^{\omega^\omega},\dots\}$$
And then we have our first fixed-point. $$\omega^{\epsilon_0}=\epsilon_0$$
Funny point in the storyline, because it's not too hard to get past this. $$\omega^{\epsilon_0+1}=\omega^{\epsilon_0}\omega=\epsilon_0\omega$$
We then diagonolize the omega out, which gives us a sum of $\epsilon_0$, which we then diagonolize one by one (and remember folks! Whenever you can, reduce the fast growing hierarchy down by turning it into repeated iterations of itself (see rule 2))
Anyways, we can then produce a higher ordinal! $$\epsilon_1=\sup\{\epsilon_0+1, \omega^{\epsilon_0+ 1}, \omega^{\omega^{\epsilon_0+ 1}}, \dots\}$$
moral of the story at this step. You diagonolize the inside part first. Then work your way out, which is how we diagonolize something like this: $$\epsilon_{\epsilon_0}=\epsilon_{\sup\{ \omega, \omega^\omega,\dots \}}$$
Well, this is something basic but I can not figure out the formal argument. Let's say we have a matrix $ M $ such that there is exactly one $ 1 $ in each column and in each row. Then $ M $ is a square matrix.
Someone please help @Xam with his problem. Please and thanks!
This is the Realm of Simply Beautiful…
This is the Realm of Simply Beautiful…