How can we prove that p(n) is true for all integers n with 1 <= n < v.size()?

AI Thread Summary
The discussion revolves around proving the proposition p(n): v.at(n) ≥ 11 + n for a sorted vector of unique integers v. For p(1), it is established that v.at(1) equals 12, which satisfies the condition since 12 is greater than or equal to 12. The challenge lies in the inductive step, where assuming p(k - 1) leads to demonstrating that p(k) holds true by leveraging the properties of sorted vectors and unique elements. The participants emphasize that since v is sorted and no two elements are the same, it follows that v.at(n) must be greater than v.at(n - 1), which aids in the inductive proof. Overall, the discussion highlights the difficulties in applying mathematical induction to this specific problem, particularly in structuring a coherent proof.
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I cannot get my head around this at all.

Suppose that v is of type Vector of Integers and we know the following:-

1. v is sorted.
2. No two items in v are the same.
3. v.at(1) is 12.

For an integer n, where 1 <= n <= v.size(), let p(n) be the following proposition: p(n): v.at(n) => 11 + n.

A. Explain why p(1) is true?
B. suppose that p(k - 1) is true (where 1 < k <= v.size()). Explain why p(k) must then be true.
C. Complete a proof that p(n) is true for all integers n with 1 <= n < v.size().

For A. I have,

In plain English the proposition means,
The n at position n in v is greater than or equal to 11 + n.

Let n = 1.

P(1): v.at(1) ≥ 11 + 1 = 12. (B)

From statement 3 which indicates v.at(1) is 12 and statement 2 which indicates no two items in v are the same, the above statement proves p(1) is true.

I'm struggling with the Inductive hypothesis (question b).
 
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Try using the facts that v is sorted and no two elements of v are the same. What does this tell you?
 
This tells me that:-

Let n = 2.

P(2): v.at(1) ≥ 11 + 2 = 13. (B)
P(3): v.at(1) ≥ 11 + 3 = 14. (B)
P(4): v.at(1) ≥ 11 + 4 = 15. (B)
P(5): v.at(1) ≥ 11 + 5 = 16. (B)
 
If they are sorted and no two are the same, doesn't this mean that v.at(n)>v.at(n-1)?
 
Yes it does.

Given that v is sorted and no item in the vector is identical then:-

v.at(n) > v.at(n - 1).

Where does p(k - 1) come into play?
 
I'm still struggling to work this out, so any help appreciated.
 
Since they are integers and v.at(n)>v.at(n-1), then v.at(n)>=v.at(n-1)+1. Can't you then work by induction?
 
What your saying makes sense to me, I understand it. I'm not sure how p(k - 1) comes into play. I fairly sure I'm learning from bad text because the book is based entirely on using mathematical induction as proof for recurrence systems. Yet this question doesn't involve a recurrence system. I've tried looking for induction learning material but nothing I've come across really helps.
 
Still struggling with this, feel as though I'm banging my head against a brick wall. I'm not sure how to put it all together into a meaningful proof.
 
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