Difference between revisions of "Proof by contrapositive"
(→Solution: oddness) |
m (wik) |
||
| Line 1: | Line 1: | ||
| − | '''Proof by contrapositive''' is a method of prove in which the contrapositive of the desired statement is proven, and thus it follows that the original statement is true. Generally, this form is only used when it is impossible to prove the original statement directly. | + | '''Proof by contrapositive''' is a method of [[proof writing|prove]] in which the [[contrapositive]] of the desired statement is proven, and thus it follows that the original statement is true. Generally, this form is only used when it is impossible to prove the original statement directly. |
==Examples== | ==Examples== | ||
===Parity=== | ===Parity=== | ||
====Problem==== | ====Problem==== | ||
| − | Show that tf <math>x</math> and <math>y</math> are two integers for which <math>x+y</math> is even, then <math>x</math> and <math>y</math> have the same parity. | + | Show that tf <math>x</math> and <math>y</math> are two integers for which <math>x+y</math> is even, then <math>x</math> and <math>y</math> have the same [[parity]]. |
====Solution==== | ====Solution==== | ||
The contrapositive of this is | The contrapositive of this is | ||
| − | + | ||
| + | :If <math>x</math> and <math>y</math> are two integers with opposite parity, then their sum must be odd. | ||
| + | |||
So we assume <math>x</math> and <math>y</math> have opposite parity. Since one of these integers is even and the other odd, there is no loss of generality to suppose <math>x</math> is even and <math>y</math> is odd. Thus, there are integers <math>k</math> and <math>m</math> for which <math>x = 2k</math> and <math>y = 2m+1</math>. Now then, we compute the sum <math>x+y = 2k + 2m + 1 = 2(k+m) + 1</math>, which is an odd integer by definition. | So we assume <math>x</math> and <math>y</math> have opposite parity. Since one of these integers is even and the other odd, there is no loss of generality to suppose <math>x</math> is even and <math>y</math> is odd. Thus, there are integers <math>k</math> and <math>m</math> for which <math>x = 2k</math> and <math>y = 2m+1</math>. Now then, we compute the sum <math>x+y = 2k + 2m + 1 = 2(k+m) + 1</math>, which is an odd integer by definition. | ||
| + | |||
===Odd Squares=== | ===Odd Squares=== | ||
====Problem==== | ====Problem==== | ||
Revision as of 21:48, 22 October 2007
Proof by contrapositive is a method of prove in which the contrapositive of the desired statement is proven, and thus it follows that the original statement is true. Generally, this form is only used when it is impossible to prove the original statement directly.
Contents
[hide]Examples
Parity
Problem
Show that tf 
and 
are two integers for which 
is even, then and have the same parity.
Solution
The contrapositive of this is
- If
and 
are two integers with opposite parity, then their sum must be odd.
So we assume and have opposite parity. Since one of these integers is even and the other odd, there is no loss of generality to suppose is even and is odd. Thus, there are integers 
and 
for which 
and 
. Now then, we compute the sum 
, which is an odd integer by definition.
Odd Squares
Problem
Show that if 
is an odd integer, then 
is odd.
Solution
Suppose is an even integer. Then there exists and integer 
such that 
. Thus 
. Since 
is an integer, is even. Therefore is not odd.
