our videos are really great, thank you very much:-) One thing I realized in this video though: Doesn't one need to include 0 into the possible values of k? Because one can only apply the proven part with 2k+1 -> 2k+3 to all n, if the value one has checked for n is included in the domain of k, which is not the case if k starts with 1, because then you start with n=3, but it is included if k start with 0.

Good explanation, perhaps also consider the following lines of working: 3^(2k+3)+7^(2k+3) = 9*3^(2k+1)+49*7^(2k+1) = (10-1)*3^(2k+1)+(50-1)*7^(2k+1) = 10*(3^(2k+1)+5*7^(2k+1))-(3^(2k+1)+7^(2k+1)) = 10*A + 10*B, where A and B are positive integers.

@darrirro

5 жыл бұрын

nice

Nice work!

Great, thanks.

You r great

I dont know how to choose n value suit with step 2? Why we can choose n=2k+3 but not another value?

@ratpete

6 жыл бұрын

Because you are trying to prove it for odd numbers. So 2k+2 is not good for you because 2*k is an even number so if you add to it another even number it will be an even number, and that's not good for you because you need an odd number. But if you write 2k+5 or 2k+anything odd, it will work but it will result in 3^5 and 7^5 or even something greater than that so it would be a lot harder to do the same thing which was presented in the video. I hope I helped and if I got anything wrong please let me know.

@wayneblackburn9645

6 жыл бұрын

You need consecutive odd values to prove that it works for all odd values. 2k+3 is the next odd value after 2k+1. Proving for 2k+5 if 2k+1 is true means you're only proving for half the odd numbers - you'd need another initial case (prove for n=1 and n=3) to prove for two sets - 1,5,9,13..... and 3,7,11,15.....

Shouldnt k be a positive integer including zero?

How are we sure that divided by ten it is an odd number?

@wayneblackburn9645

6 жыл бұрын

You don't know that but that that hasn't been said anywhere as far as I can see. The only thing that's stated to be odd is n which is why it can be represented as 2k+1 where k is any whole number.

Sir I want to know how is e^πi=-1

@dominicgagliardi5864

6 жыл бұрын

check out these channels: Numberphile and 3blue1brown

@lightningkiki6090

2 жыл бұрын

Write the maclaurin series expansions of cos(theta), sin(theta) and e^x Replace x with i(theta) and multiply sin(theta) expansion by i You should find that e^i(theta) is equal to cos(theta)+isin(theta). This is called Euler’s relation. If you substitute in theta= pi into this and rearrange you will get e^ipi= -1 (as long as you remembered to use radians that is)

How about - proving it's true for any positive integer, which by default proves it is true for positive odd integers? Easier?

2 жыл бұрын

But it's not true for n=2 for example.

The second part was nonsense