Let p be an odd integer.
for n=1 which is odd.
Suppose is odd.
Now which by part a is odd.
So the answer on the sheet was a little off.
hi the link to this question is here http://i223.photobucket.com/albums/d...f/HPIM0535.jpg
Question 8. I think I was OK in the first part with proving 2n(2n+1) is always odd.
However I am lost with part b p^n. I'm not sure if the answer written on the scan is correct.
Thanks
No, it's not. You're not being asked to prove something about the special case of two consecutive odd numbers, but rather about ANY pair of odd numbers. And I gave the answer just above.
Suppose p is odd. Show p^n is odd. Induction on n:
n=1. Then p^n = p is odd.
Suppose p^n is odd. Show p^(n+1) is odd.
p^(n+1) = (p^n)*p. But we have p^n is odd, and p is odd, and the product of two odd numbers is odd, so p^(n+1) is odd.
This is not the product of 2 odd numbers as 2n is even.
Let be odd.
Then is also odd.
P(k)
P(k+1)
Show that P(k) being true will cause P(k+1) to be true
Proof
?
If P(k) is true, then is odd,
therefore since 2n is even and odd+even=odd, so P(k+1) is true if P(k) is.
Then test for an initial value, such as n=1.
There's no need to use induction for part (a) but it can be used nevertheless.
Part (b)
p=odd
P(k)
P(k+1)
Try to show that P(k) being true causes P(k+1) to be true
Proof
if P(k) is true
Test for an initial value