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...

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.

I have seen your sister online a few times in the last year, she seems OK. Ms Sue, Writeteacher, and myself are very proud of you.

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 :)

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?

For n+1 I got the answer 3...

Sure! I'd be happy to help you with your homework question.

To prove that the output always ends up as 1, we can go through the steps using a variable 'n' as the starting number and see if the result is consistently 1.

Let's go through each step one by one and see how it affects the value of the variable 'n':

1. Think of a number between 1 and 10: At this point, the value of 'n' is the number you chose.

2. Add 1: The value of 'n' becomes 'n + 1'.

3. Double the result: The value of 'n' becomes '2*(n + 1)'.

4. Add 3: The value of 'n' becomes '2*(n + 1) + 3'.

5. Subtract 4: The value of 'n' becomes '2*(n + 1) + 3 - 4'.

6. Add 5: The value of 'n' becomes '2*(n + 1) + 3 - 4 + 5'.

7. Halve the result: The value of 'n' becomes '(2*(n + 1) + 3 - 4 + 5)/2'.

8. Add 6: The value of 'n' becomes '(2*(n + 1) + 3 - 4 + 5)/2 + 6'.

9. Subtract 7: The value of 'n' becomes '(2*(n + 1) + 3 - 4 + 5)/2 + 6 - 7'.

10. Add 8: The value of 'n' becomes '(2*(n + 1) + 3 - 4 + 5)/2 + 6 - 7 + 8'.

11. Subtract 9: The value of 'n' becomes '(2*(n + 1) + 3 - 4 + 5)/2 + 6 - 7 + 8 - 9'.

12. Subtract the number you first thought of: The value of 'n' becomes '(2*(n + 1) + 3 - 4 + 5)/2 + 6 - 7 + 8 - 9 - n'.

Now, let's simplify this expression step by step:

- If we simplify '(2*(n + 1) + 3 - 4 + 5)/2' further, we get '(2n + 8)/2' which simplifies to 'n + 4'.
- So, the expression becomes '(n + 4) + 6 - 7 + 8 - 9 - n'.
- Simplifying this further, we get 'n + 2'.

From this simplified expression, it is clear that the value of 'n' cancels out, and the output expression always becomes '2'.

Therefore, the equation is incorrect and the output does not always end up as 1, regardless of the starting number.

In summary, by analyzing the expression and simplifying it, we can see that the output always becomes 'n + 2', not 1.