Question 1173524
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They are illustrating proof by induction<br>

(1)  8-3  is divisible by 5  (should not have ^1)
(2)  8k-3k  is divisible by 5  (this is the hypothesis)
(3) Let n=k+1; then...
8(k+1) - 3(k+1) 
= 8k+8 - (3k+3) 
= 8k+8 -3k-3
= (8k-3k)+5 

By hypothesis, 8k-3k is divisible by 5  (that was the n=k case)

so  8k-3k+5 is also divisible by 5 (if  F is divisible by 5, so is F+5)


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This part:
"III. we need to show that the statement is also true for the next integers n=k+1, i.e. 8k+1=3k+1 is divisible by 5.
consider: 8k+1-3k+1=8k*8-3k*3"

should be:
"III. we need to show that the statement is also true for the next integers n=k+1,  i.e. 8(k+1)-3(k+1) is divisible by 5.
Consider:  8(k+1)-3(k+1) = 8k+8-3k-3"  


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While it seems they are trying to show how proof by induction works, note the original problem "show 8n-3n is divisible by 5" is much more easily proven directly:   8n-3n = 5n and 5*(any integer) is obviously divisible by 5.  

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IMO, a better problem is to show  1+2+...+n  = (1/2)(n+1)(n) by induction, which is also easily shown by direct proof but not quite as obvious.

Prove  S = 1+2+3+...+n = (1/2)(n+1)(n)  by induction 

n=1:  S=1   and  (1/2)(1+1)(1) = 2/2 = 1   (holds for n=1, this is the base case)
Assume true for n=k  (*)   (the hypothesis)

Let n=k+1:   S = [ 1+2+...+k ] + (k+1) = [ (1/2)(k+1)(k) ] + (k+1)
where we applied the hypothesis (*) for bracketed [ ] items.

Factor out k+1:  S = (k+1) ((1/2)(k) + 1)

         since  (1/2)k + 1 =  k/2 + 2/2 =  (k+2)/2  we end up with

     S = (k+1)(k+2)/2 =  (1/2)(k+1)(k+2)  
( if you write this in terms of n you have S = (1/2)(n)(n+1) )  

thus we have shown the hypothesis holds true for n=k+1 and the proof is complete.


...and if you wish, by direct proof:

   write the su              S = 1 +  2 +    3 +...+ (n-1) + n
   also write it             S = n +(n-1)+(n-2) +...  + 2  + 1
                            -----------------------------------
   add both eqns            2S = (n+1)+(n+1)+(n+1)+...+(n+1)+(n+1)

The RHS has n terms of n+1, that's just n*(n+1):
                            2S = n*(n+1)
                     
              and finally    S = (1/2)*n*(n+1)