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/sci/ - Science & Math

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File: 28 KB, 488x463, 1520114077531.png [View same] [iqdb] [saucenao] [google]
9561315 No.9561315 [Reply] [Original]

Guise, using Peano's axioms I want to prove that

m · S(n) = nm + m
where S(n) is the succesor of n.

So, if m = 0
0 · S(n) = n·0 + 0
0 = 0 + 0 so it's true (I proved beforehand that 0·n = n·0)

I must check that
S(m)·S(n) = n·S(m) + S(m)

Since multiplication is defined by
S(x) y = xy + y
and addition is defined by
S(x) + y = S(x + y) then

(Left hand)
S(m)·S(n) = mS(n) + S(n)
= nm + m + S(n) (by Hypoteshis)
= nm + S(m + n)
= S(nm + (m + n) )

(Right hand)
n·S(m) + S(m) = mn + n + S(m) (by hypothesis)

= mn + S(n + m)
= S(mn + (n + m) )
=S(mn + (m + n) )

So, if L.H equals R.H
S(nm + (m + n) ) = S(mn + (m + n) )
it means
nm + (m + n) = mn + (m + n)
Which implies
nm = mn

And substituting back there we finally get
S(mn + (m + n) ) = S(mn + (m + n) )

Is that correct?
Also, did I just proved that nm = mn implicitly?

>> No.9561338

>>9561315
Your proof sucks.

>> No.9561396

>>9561315
Can you write it in Coq? This is very difficult to read

>> No.9562119

>>9561396
I'll rewrite it properly, wait

>> No.9562263
File: 36 KB, 484x797, tits.png [View same] [iqdb] [saucenao] [google]
9562263

can you help me now please?

>> No.9563565

>>9562263
whats with the colored text shit

>> No.9563633
File: 28 KB, 680x383, 075.jpg [View same] [iqdb] [saucenao] [google]
9563633

>>9561315
>PBS infinite series

>> No.9563686

>>9561315
>Is that correct?
No.

After manipulating that left hand side and right hand side, what *remains to be proven* is that S(nm + (m + n)) = S(mn + (m + n)). That is not something you have, that's something you need to prove. Which means you cannot conclude that nm = mn.

>> No.9563709

>>9563686
you're right, in the image I uploaded I corrected that and now I'm stuck there. Can you help me??

>> No.9563774

Don't do left hand side and right hand side shit with induction. You just need to prove the lemmas that multiplication and addition are commutative (by induction).

>> No.9563783

>>9563774

That's why I'm trying to do. Prove that nm = mn, but that means I must prove that S(n)m = mS(n), and that's exactly what I'm stuck at.

>> No.9563793

>>9563783
that's why I'm trying to prove that mS(n) = nm + m, and thus S(n)m = mS(n)