How to prove by mathematical induction?by James2 Tags: formulas, induction, logic, mathematics, proof 

#1
Feb313, 10:22 PM

P: 35

How do I prove a formula/rule or something by mathematical induction? Please give me a few examples or resources and explain it as best you can because I think I'm messing up some how.




#2
Feb313, 10:34 PM

P: 771

What are you not getting, exactly? If you just don't know what induction is, surely a google search would be faster than starting a new thread.




#3
Feb313, 10:39 PM

P: 35

Everything I read confuses me, it tells me to do something different everytime...




#4
Feb313, 10:40 PM

P: 771

How to prove by mathematical induction? 



#5
Feb313, 10:56 PM

P: 35

I have an equation, (5n + 2) = 2[(5/2)n + 1] I know this is true from the basis step. Then I asume n = k now I must prove n = k + 1. So, (5k + 2) = 2[(5/2)k + 1]
Alright then, I try to substitute k + 1 in and add it or something so I get... 2[(5/2)k + 1] + [5(k + 1) + 2] = 2 [(5/2)(k + 1) + 1] Simplifying, I get 10k + 2 + 5k + 5 + 2 = 10(k + 1) + 2 And finallly, 15k + 9 =/= 10k + 12 SO.... whaaaat? What happened here? 



#6
Feb413, 12:42 AM

P: 22

(5n + 2) = 2[(5/2)n + 1]
n = 0: 5*0 + 2 = 2[(5/2)*0 + 1] The case is true for 0. Suppose the case is true for n = k. Now we can use (5k + 2) = 2[(5/2)k + 1]. n = k + 1: 5(k + 1) + 2 = (5k + 2) + 5 = 2[(5/2)k + 1] + 5 = 2[(5/2)k + 1 + 5/2] = 2[(5/2)(k+1) + 1] The case n = k + 1 follows from the case n = k. With case n = 0 true the equation therefore works for all nonnegative integers. 


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