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You are not logged in. #1 20060228 15:48:15
Mathematical InductionMI # 1 Character is who you are when no one is looking. #2 20060228 17:11:19
Re: Mathematical Inductionk.k!= (k+1)k!  k! = (k+1)!  k! Last edited by krassi_holmz (20060228 17:14:10) IPBLE: Increasing Performance By Lowering Expectations. #3 20060302 03:39:10
Re: Mathematical InductionI have no idea what those numbers say or mean.Probably because we haven't learn't about whatever you're talking about yet. Presenting the Prinny dance. Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! #4 20060302 03:42:20
Re: Mathematical InductionIt's never early to learn something: Last edited by krassi_holmz (20060302 04:00:39) IPBLE: Increasing Performance By Lowering Expectations. #5 20060302 03:55:50
Re: Mathematical Inductionsorry but I haven't even learnt what an integer is. Presenting the Prinny dance. Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! #6 20060302 04:00:53
Re: Mathematical InductionOK. IPBLE: Increasing Performance By Lowering Expectations. #7 20060302 04:03:04
Re: Mathematical Inductionwhen did you learn? Presenting the Prinny dance. Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! #8 20060302 04:09:12
Re: Mathematical InductionI'm from Bulgaia. I can't tell you. IPBLE: Increasing Performance By Lowering Expectations. #9 20060302 05:01:55
Re: Mathematical InductionThe differences between integers, rationals, and reals don't really have to be well defined (bad pun...) to understand them. Just think of an integer as any whole number, without a decimal or fraction, a ration is anything that can be put in the form a/b (where a and b are integers), and real includes all the numbers without a complex part (i). Last edited by Ricky (20060302 05:02:27) "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #10 20060302 15:07:14
Re: Mathematical Inductionkrassi_holmz, I am not able to understand your proof. If no other member posts the solution, I shall do it before posting the next problem. Character is who you are when no one is looking. #11 20060302 16:21:32
Re: Mathematical InductionI'm expressing kk! of the form f(k+1)f(k) so the sum When I do this all fs between 2 and n reduct as you see because f(k) thakes part in f(k+1)f(k) with sign "" and in f(k)f(k1) whit sign "+". So there left only f(1) and +f(n+1): Now it's not hard to prove that kk!= (k+1)!k!: (k+1)!k!= ((k+1)k!)k!=k!(k+11)=kk! so for every k kk!= (k+1)!k!=f(k+1)f(k),where f(x)=x! So the sum: , which have to be proven. Last edited by krassi_holmz (20060302 16:24:50) IPBLE: Increasing Performance By Lowering Expectations. #12 20060923 12:48:14
Re: Mathematical InductionCAn you help me with a question about mathematical induction? #13 20110301 18:38:07
Re: Mathematical Induction
You wrote n=1 instead of n The statement is true for some p. Maths!...... 