Proof Accuracy

[toughcookie723]

Would really appreciate if someone took a look at whether this proof is correct:

The sum of n non-negative integers, each less than n, is less than n^2.

0+1+2+3+......+(n-1)+(n-1)< n^2--proved it's true

Therefore, if true for n=k, then true for n=k+1.

0+1+2+3+...+(k-1)+(k-1) < k^2
+2k+1 +2k+1
____________________________
0+1+2+3+...+(k-1)+(k-1)+2k+1 < k^2+2k+1
0+1+2+....+2k-2+2k+1 < k^2+2k+1
0+1+2+......+(2k-1)+2k< k^2+2k+1
0+1+2+....+2k < (k+1)^2
0+1+2+......+(k+1)-1+ (k+1)-1< (k+1)^2


Thank you!

 

[pka]

Would really appreciate if someone took a look at whether this proof is correct:
The sum of n non-negative integers, each less than n, is less than n^2.
0+1+2+3+......+(n-1)+(n-1)< n^2--proved it's true

Therefore, if true for n=k, then true for n=k+1.
0+1+2+3+...+(k-1)+(k-1) < k^2
+2k+1 +2k+1___________________________
0+1+2+3+...+(k-1)+(k-1)+2k+1 < k^2+2k+1
0+1+2+....+2k-2+2k+1 < k^2+2k+1
0+1+2+......+(2k-1)+2k< k^2+2k+1
0+1+2+....+2k < (k+1)^2
0+1+2+......+(k+1)-1+ (k+1)-1< (k+1)^2

Quite frankly, I follow none of the above.
We have n non-negative such that \(\displaystyle 0\le a_k< n,~1\le k\le n\).

Surely we know \(\displaystyle \sum\limits_{k = 1}^n {a_k } \leqslant \sum\limits_{k = 1}^n {\left( {n - 1} \right)} = n\left( {n - 1} \right) < ?\)

 

[Someone2841]

Would really appreciate if someone took a look at whether this proof is correct:

The sum of n non-negative integers, each less than n, is less than n^2.

0+1+2+3+......+(n-1)+(n-1)< n^2--proved it's true

Therefore, if true for n=k, then true for n=k+1.

0+1+2+3+...+(k-1)+(k-1) < k^2
+2k+1 +2k+1
____________________________
0+1+2+3+...+(k-1)+(k-1)+2k+1 < k^2+2k+1
0+1+2+....+2k-2+2k+1 < k^2+2k+1
0+1+2+......+(2k-1)+2k< k^2+2k+1
0+1+2+....+2k < (k+1)^2
0+1+2+......+(k+1)-1+ (k+1)-1< (k+1)^2


Thank you!

There's no way I can vouch for the validity of your proof, since you started by assuming \(\displaystyle 0+1+2+3+\ldots+(n-1)+(n-1)< n^2\) to be true.

Rather than starting with induction, I would start with this equality: \(\displaystyle \underset{i=0}{\overset{n-1}{\sum}} i=\frac{1}{2} \left(n^2-n\right)\) (can be proven for all positive integers with induction). From there, you just need to show that \(\displaystyle \frac{1}{2} \left(n^2-n\right) < n^2\) for all \(\displaystyle n>0\).

 

[SAMUELK]

Would really appreciate if someone took a look at whether this proof is correct:

The sum of n non-negative integers, each less than n, is less than n^2.

0+1+2+3+......+(n-1)+(n-1)< n^2--proved it's true

Therefore, if true for n=k, then true for n=k+1.

0+1+2+3+...+(k-1)+(k-1) < k^2
+2k+1 +2k+1
____________________________
0+1+2+3+...+(k-1)+(k-1)+2k+1 < k^2+2k+1
0+1+2+....+2k-2+2k+1 < k^2+2k+1
0+1+2+......+(2k-1)+2k< k^2+2k+1
0+1+2+....+2k < (k+1)^2
0+1+2+......+(k+1)-1+ (k+1)-1< (k+1)^2


Thank you!

Of course your proof is INCORRECT! Instead of proving that

The sum of n non-negative integers, each less than n, is less than n^2.

you are proving that

The sum of n non-negative CONSECUTIVE integers, each less than n, is less than n^2.

--- not at all the same thing.

Instead, take as your 'k-case':

A[1] + A[2] + ... + A[k] < k^2, with each A[j] < n

Now try to prove that

B[1] + ... B[k] + B[k + 1] < (k + 1)^2, with each B[j] < n + 1.

I think that if you just:

1. Write B[j] = A[j] + 1.

2. Show that each A[j] < n, (not too hard)

3. Show that:

B[1] + ... + B[k] + B[k + 1] =

A[1] + ... + A[k] + A[k + 1] +
1 + ... + 1 + 1 =

A[1] + ... + A[k] + A[k + 1] + k + 1 < (by hyp)

k^2 + A[k + 1] + k + 1 <

k^2 + k + k + 1 = (k + 1)^2






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