For 1.
P(k)
P(k+1)
Show that "if" P(k) is valid, then P(k+1) "must therefore" also be valid.
Hence, write P(k+1) in terms of P(k).
Proof
If P(k) is valid, then
If , then P(k+1) will be valid.
If
then
It can be seen that this is true since
hence, you need to examine the case for
For 2.
You want to show that this inequality holds true for any level of nesting.
Suppose the above inequality is at a nesting depth of or
P(k)
for a nesting depth of k.
Add to both sides and take the square root to get a depth of
The above has a nesting depth of
P(k) will definately be valid if
If this is so, then P(k) is always true.
Squaring both sides...
??
It's now a simple matter to verify this.
Finally, test for or
?
?
which is certainly true as
No.4 was a bit more challenging.
Writing it in the way we normally use "k",
P(k)
P(k+1)
We want to show that if P(k) is valid, then it will cause P(k+1) to be valid also.
Hence we try to write P(k+1) so that it contains the P(k) expression.
Then we can draw the conclusion.
Proof
Rewriting P(k+1)..
Squaring both sides,
Subtracting from both sides,
Hence, our rewritten P(k+1) is
Rewriting P(k) by squaring both sides, to examine P(k+1)...
Examining the above rewritten P(k+1) and P(k) propositions,
it's easy to see that the RHS of the P(k+1) expression is greater than the RHS of the P(k) expression,
while the LHS is the same for both.
Therefore, if P(k) is valid, P(k+1) will as a result, also be valid.
Hence you only need to test for or