- #1

mr_coffee

- 1,629

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Hello everyone.

I"m trying to break down this problem, the answers are given I just want to make sure I'm understanding it right.

http://img204.imageshack.us/img204/9936/lastscannd1.jpg

OKay here is my reasoning:

For the first one, which is invalid.

they say For n >= 3, assume that P(k) is true for all k >=1 such that k <= n. So k <= 3. We will porve that P(n) is true.

Assumes P(1), P(2), P(3)

Is it invalid because P(3) wasn't proven true in the beginning, it ony said:

P(1) and P(2) are true.For the 2nd, which is valid.

For n >= 2, assume that P(k) is true for all k >= 1 such that k < = n, so

k <= 2. We will prove that P(n+1) is true.

Proves: P(1), P(2), and P(2+1) = P(3).

Assumes P(1) and P(2) are true, which it is, from the first pargraph, so that must be valid.

For the 3rd, which is also valid:

For n > 3, assume that P(k-1) is true for all k >= 2 such that k < n. We will prove that P(n-1) is true.

Proves P(1),P(2), P(3), and P(4)

assumes P(1), P(2).

So k = 2, 3, 4

n > 4, so n = 4

assumes P(k-1) so we have:

assumes P(3), P(2), P(1)

For the 4th one, which is invalid:

FOr n > 3, assume that P(k) is true for all k >= 1 such that k < n. We will prove that P(n) is true.

Prove: P(n), and n > 3, so it will prove P(4), P(5), P(6)

assume P(k) is true for all k >=1 such that k < n.

So, k = 1, 2, 3

assume P(1), P(2), P(3),

we are assuming P(3) but there is no proving of P(3) becuase the proving starts at P(4) is that why this is invalid?For the 5th one, which is invalid

For n >=3, assume that P(K) is true for all k >= 0 such that k < n. We will prove that P(n) is true.

n = 3, 4, 5...

k = 0, 1, 2, 3, 4, 5...

assumes: P(0),P(1),P(2),P(3)

proves: P(3), P(4), P(5)

is this invalid because again ur assuming P(3), when P(3) wasn't proven to be true in the beginning paragraph?For number 6, which is valid:

For n >= 1, assume that P(k) is true for all k >=1 such that k <= 2n. We will prove that P(2n+1) and P(2n+2) are true.Proves n = 1: P(3) and P(4)

assumes n = 1: P(1), P(2)

Proes n = 2: P(5) and P(6)

assumes n = 2: P(1),P(2),:(3),P(4)

I see this is convering all the spots but for n = 1, why does it stop at P(2), and for n = 2, why does it stop at P(4)?Thanks!

I"m trying to break down this problem, the answers are given I just want to make sure I'm understanding it right.

http://img204.imageshack.us/img204/9936/lastscannd1.jpg

OKay here is my reasoning:

For the first one, which is invalid.

they say For n >= 3, assume that P(k) is true for all k >=1 such that k <= n. So k <= 3. We will porve that P(n) is true.

Assumes P(1), P(2), P(3)

Is it invalid because P(3) wasn't proven true in the beginning, it ony said:

P(1) and P(2) are true.For the 2nd, which is valid.

For n >= 2, assume that P(k) is true for all k >= 1 such that k < = n, so

k <= 2. We will prove that P(n+1) is true.

Proves: P(1), P(2), and P(2+1) = P(3).

Assumes P(1) and P(2) are true, which it is, from the first pargraph, so that must be valid.

For the 3rd, which is also valid:

For n > 3, assume that P(k-1) is true for all k >= 2 such that k < n. We will prove that P(n-1) is true.

Proves P(1),P(2), P(3), and P(4)

assumes P(1), P(2).

So k = 2, 3, 4

n > 4, so n = 4

assumes P(k-1) so we have:

assumes P(3), P(2), P(1)

**But what confusees me is, if its assuming P(1),P(2), and P(3), and P(4),**

the beginning paragraph didn't say P(3) and P(4) were true, so why is this valid?the beginning paragraph didn't say P(3) and P(4) were true, so why is this valid?

For the 4th one, which is invalid:

FOr n > 3, assume that P(k) is true for all k >= 1 such that k < n. We will prove that P(n) is true.

Prove: P(n), and n > 3, so it will prove P(4), P(5), P(6)

assume P(k) is true for all k >=1 such that k < n.

So, k = 1, 2, 3

assume P(1), P(2), P(3),

we are assuming P(3) but there is no proving of P(3) becuase the proving starts at P(4) is that why this is invalid?For the 5th one, which is invalid

For n >=3, assume that P(K) is true for all k >= 0 such that k < n. We will prove that P(n) is true.

n = 3, 4, 5...

k = 0, 1, 2, 3, 4, 5...

assumes: P(0),P(1),P(2),P(3)

proves: P(3), P(4), P(5)

is this invalid because again ur assuming P(3), when P(3) wasn't proven to be true in the beginning paragraph?For number 6, which is valid:

For n >= 1, assume that P(k) is true for all k >=1 such that k <= 2n. We will prove that P(2n+1) and P(2n+2) are true.Proves n = 1: P(3) and P(4)

assumes n = 1: P(1), P(2)

Proes n = 2: P(5) and P(6)

assumes n = 2: P(1),P(2),:(3),P(4)

I see this is convering all the spots but for n = 1, why does it stop at P(2), and for n = 2, why does it stop at P(4)?Thanks!

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