Hello Mario,

Sorry if this is too trivial of a question. Where does this statement [ z
 < 1 -> A. x e. ZZ ( sqrt ` 2 ) =/= ( x / z ) ] come from? I understand 
it's associated to ( n = 1 ) case for strong induction, but I am not sure 
what makes it true. 

Also, is there a way to know what book this theorem was referenced from? I 
apologize since these questions aren't really related to Metamath but math 
in general. Let me know if I need to stop posting these kind of questions 
here. Appreciate your time.

-Malik.

On Sunday, November 20, 2022 at 9:53:03 PM UTC-5 [email protected] wrote:

> On Sun, Nov 20, 2022 at 9:24 PM M Malik <[email protected]> wrote:
>
>> Hello Metamath peeps,
>>
>> I am trying to learn advanced math with assistance of Metamath. I 
>> understand a lot of elementary theorems that are in the database. However, 
>> I thought I would jump to (sqrt2irr -Irrationality of square root of 2), 
>> but its proof isn't really making sense to me. 
>>
>> I am specifically confused about this wff [ For all x e. Z (sqrt2 =/= 
>> x/z) ]. Is this something to be assumed true from the start?
>>
>
> No, it (more or less) is proved by complete induction on z (which is to 
> say, induction on y e. NN such that x < y) at step 78 of 
> https://us.metamath.org/mpeuni/sqrt2irr.html .
>   
>
>> Also, in general, is it recommended to download the mmj2 program or the 
>> content on the Metamath website sufficient enough?
>>
>
> The content of the website should be sufficient for viewing proofs, and 
> indeed is somewhat better than the mmj2 display for many purposes. mmj2 is 
> more if you are interested in writing a proof yourself.
>
> Mario
>

-- 
You received this message because you are subscribed to the Google Groups 
"Metamath" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to [email protected].
To view this discussion on the web visit 
https://groups.google.com/d/msgid/metamath/b31cd4ce-0dff-496f-9892-b1a714987710n%40googlegroups.com.

Reply via email to