211Lecture4: Difference between revisions
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Latest revision as of 22:35, 8 March 2008
pavel@cs.auckland.ac.nz
576(CS Building)
1pm-2pm Tuesday = Office hours
Questions
Example 1
If n, m are odd numbers, then n.m is odd.
Proof
There are Integer numbers K and L such that
n=2K+1, m=2L+1
n.m = (2K+1).(2L+1) = 4KL+2(K+L)+1 - this is an odd number
Example 2
If p,q are rational numbers, then p-q is rational.
Proof
p=n/m, where n,m are Integers and m != 0.
q=l/k where k, and l are Integers and k != 0.
p-q = n/m - l/k = (nk-lm)/mk = this is a rational number.
- Example 1 and Example 2 are examples of direct proofs.
Example 3
If p is rational, and q is an irrational number, then p+q is irrational.
Proof
Assume that p+q is rational. Let l=p+q.
q = t - p is a rational number. By hypothesis, it is Irrational
- This example is done using proof by contradiction. QED
Example 4
If 3/(n^2) then 3/n, where n is an Integer.
Proof
Assume that 3/(n^2)
n = 3k + r, where 0<=r<=3 (r is the remainder) r = 0,1,2
So r can be 1 or 2
- let r = 1
- n^2 = (3k+1)^2 = y*x^2 + b*k+1 hence 3/(n^2)
- let r = 2
- n^2 = (3k+2)^2 = 9k^2 + 12k + 4 =
- 3(3k^2 + 4k +1) + 1 thus 3/(n^2) QED
Example 5
sqrt(3) is Irrational
Proof
Assume that sqrt(3) = n/m, where n,m are Integers, m != 0 , and n,m do not have common factors.
Therefore 3 = n/m => 3m^2 = n^2 Then 3/(n^2) => 3/n.
There is k such that n = 3k. 3m^2 = 9k^2 => m^2=3k^2
So 3/m - then 3/m
If sqrt(3) = n/m then 3/n and 3/m n=3n', m=3m'
sqrt(3) = n'/m'