Mathematical Induction - Divisibility Tests (1) | ExamSolutions

Here I look at using proof by mathematical induction for divisibility tests.
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[@ExamSolutions_Maths]

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[@soccerzone9945]

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[@rosalynmaepaligutan6986]

how'd you get 8 in f(k+1)

[@ExamSolutions_Maths]

in 5:02 , If you expand the 3^2 in the front, you will get 9(3^2k) - 3^2k which equals to 8(3^2k)

[@Rachel-rp7go]

how did u get 3squre{3^2K}-3^2K

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[@suzukiliciously]

How does he get to the conclusion that it is 8(3^2k) - my question goes as to how it is 8, in particular, and what happens to the - 3^2k? Cheers guys, I'm a total rookie who just need to understand how the algebraic expression is simplified?!?

[@TadiwaChinembiri-op2wt]

I'm confused as well

[@TizitaDemiss]

3^2k(3^2-1) 3^2K(8)

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[@ExamSolutions_Maths]

cool.

[@TanakaMusasira-lf1fl]

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[@ELLIPTICALWR]

how would you use proof by induction to prove a function is not divisible by a real number, like n^3 +2 for example is not divisible by 8

[@oatmilkpeek1291]

uhh what year group is this for...

[@mattagamer98]

Thank you!

[@ExamSolutions_Maths]

You're welcome Matti! All the best

[@hendrickmafane5908]

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[@billywalshpatricianpresent3278]

great method. thanks. would this method work for something such as this....... prove 2^(3n) - 3^n is divisible by 5 for all natural numbers

[@ExamSolutions_Maths]

Yes it should

[@billywalshpatricianpresent3278]

@@ExamSolutions_Maths iv gotten as far as f(k+1) - f(k)= 7(2^3k) -2(3^k) ...........but how do i know this is also divisible by 5. thanks in advance

[@ExamSolutions_Maths]

Sorry for the delay in replying but a lot has been going on over the last few days. You may have it sorted by now but this is tricky. Now we know that f(k)=2^3k - 3^k and it is divisible by 5 (assumption) so rearranging 2^3k = f(k) + 3^k Sub this into your last statement f(k+1) - f(k)= 7(2^3k) -2(3^k) to give f(k+1) - f(k)= 7[f(k) + 3^k] -2(3^k) so f(k+1) - f(k)= 7 f(k) + 7(3^k) -2(3^k) which gives f(k+1) - f(k) = 7f(k) + 5(3^k) Now add f(k) to both sides so f(k+1)=8f(k) + 5(3^k) Now f(k) is assumed to be divisible by 5 so f(k)=5a where a is a positive integer so f(k+1)=40a + 5(3^k) so f(k+1)=5[8a +3^k] Which is a multiple of 5 so hence divisible by 5 as 8a+5(3^k) is an integer I hope that answers your question.

[@billywalshpatricianpresent3278]

@@ExamSolutions_Maths after 3 days of messing around with it i finally got it sorted. thanks for your reply though. great videos and a very nice way of doing the divisibility proofs.

[@gilbertvirgo5672]

Are you using a Blue Yeti microphone?

[@ExamSolutions_Maths]

Yes

[@2minutesquickvideos588]

Can you please explain why we don't add the n= k+1 to n=k? I am a little confused on the part where you did f(k+1)-f(k) and why you did that. I understand the rest but if you can just explain that part then all the gaps will be filled :) . Thanks

[@ExamSolutions_Maths]

Did you watch all the video? At the beginning I said this (and this applies to all divisibility tests) was handled differently from many of the other forms of proof by induction by this method of doing f(k+1)-f(k). We show that f(k+1)-f(k) = some expression and then f(k+1)=the expression +f(k). All we need to show is the expression is divisible by 8 and we know that f(k) is divisible by 8 so if both terms are divisible by 8 then f(k+1) must be divisible by 8.

[@hemodiagnostikapardes8370]

@@ExamSolutions_Maths You are not answering , You are just telling what You did

[@amythestgemini1902]

is this pre-algebrq, algebra or calculus?

[@ffaiq]

algebra I guess

[@sophia96660]

Can anyone tell me how 3^2.3^2k - 3^2k = 8(3^2k)??? Please guide me about that if anyone is reading my comment, I'm stuck here 🥺🥺

[@ketchupjunge2826]

Okay first lets simplify 3^2 which is 9 so we have: 9(3^2k)-3^2k Next we rewrite the statement like this: 9(3^2k) + (-1x 3^2k) which is the same as 9(3^2k)-3^2k we just multiplied -1 times 3^2k which is the same as subtracting it Now we can factor out 3^2k like this: 3^2k(9-1) The last step is to calculate 9-1 so we get 3^2k(8) which is the same as 8(3^2k) I hope that helps took me a while to figure this out

[@user-rw2uz2ix2q]

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[@killianversoza7101]

I miss quadratic equation...

[@Kebab_with_extra_garlic_mayo]

why did you add f(k) I do the same method but I dont add f(k) and i make F(k+1) = 8A and then i Factor

[@ExamSolutions_Maths]

I added f(k) so that f(k+1) was left on the LHS and on the RHS I had two term which contained a factor of 8. Remember that I assumed that f(k) was divisible by 8. Show me the method you used and I can then see if it is a fair argument.

[@retardedproducer4949]

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[@NaomiManda-um8rj]

Where does that 8 come from cause 3² is 9

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