your procedure is correct. If you prove that with any n, the result is one, you have it. Some would also like you to do then do the same for the number (n+m), than that is icing on the cake, we used to just do it for n+1 as induction.
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Hello all! I had a quick homework question:
I have to prove that the output of this always ends up as 1, regardless of the starting number:
*Think of a number between 1 and 10. Add 1; double the result; add 3; subtract 4; add 5; halve the result; add 6; subtract 7; add 8; subtract 9; subtract the number you first thought of.*
I am not allowed to use specific examples as a proof.
However, I decided to make 'n' my starting number and I went through and did all of the operations; my end result was still 1. I think this would count as a proof as long as I specify what 'n' is.
I believe that 'n' can be all real numbers, but I am not 100% sure.
I hope someone can help me with this...
4 answers
I followed the same steps with n+1, but somehow I did not end up with 1...did I do something wrong?
Thank you guys so much! It is nice to check in every once in a while :)
Thank you guys so much! It is nice to check in every once in a while :)
Think of a number between 1 and 10: n
Add 1; n+1
double the result; 2(n+1) = 2n+2
add 3; 2n+2+3 = 2n+5
subtract 4; 2n+1
add 5; 2n+6
halve the result; (2n+6)/2 = n+3
add 6; subtract 7; add 8; subtract 9; n+1
subtract the number you first thought of: 1
Don't see any induction required. It works for any n. What happens when you tried n+1?
Add 1; n+1
double the result; 2(n+1) = 2n+2
add 3; 2n+2+3 = 2n+5
subtract 4; 2n+1
add 5; 2n+6
halve the result; (2n+6)/2 = n+3
add 6; subtract 7; add 8; subtract 9; n+1
subtract the number you first thought of: 1
Don't see any induction required. It works for any n. What happens when you tried n+1?
For n+1 I got the answer 3...
0 answers