Editing Euclidean algorithm (section)

=== Principal ideals and related problems ===
Bézout's identity provides yet another definition of the greatest common divisor {{math|''g''}} of two numbers {{math|''a''}} and {{math|''b''}}.<ref name="Leveque_p33" >{{Harvnb|LeVeque|1996|p=33}}</ref> Consider the set of all numbers {{math|''ua'' + ''vb''}}, where {{math|''u''}} and {{math|''v''}} are any two integers. Since {{math|''a''}} and {{math|''b''}} are both divisible by {{math|''g''}}, every number in the set is divisible by {{math|''g''}}. In other words, every number of the set is an integer multiple of {{math|''g''}}. This is true for every common divisor of {{math|''a''}} and {{math|''b''}}. However, unlike other common divisors, the greatest common divisor is a member of the set; by Bézout's identity, choosing {{math|1=''u'' = ''s''}} and {{math|1=''v'' = ''t''}} gives {{math|''g''}}. A smaller common divisor cannot be a member of the set, since every member of the set must be divisible by {{math|''g''}}. Conversely, any multiple {{math|''m''}} of {{math|''g''}} can be obtained by choosing {{math|1=''u'' = ''ms''}} and {{math|1=''v'' = ''mt''}}, where {{math|''s''}} and {{math|''t''}} are the integers of Bézout's identity. This may be seen by multiplying Bézout's identity by ''m'',
: {{math|1=''mg'' = ''msa'' + ''mtb''}}.

Therefore, the set of all numbers {{math|''ua'' + ''vb''}} is equivalent to the set of multiples {{math|''m''}} of {{math|''g''}}. In other words, the set of all possible sums of integer multiples of two numbers ({{math|''a''}} and {{math|''b''}}) is equivalent to the set of multiples of {{math|gcd(''a'', ''b'')}}. The GCD is said to be the generator of the [[ideal (ring theory)|ideal]] of {{math|''a''}} and {{math|''b''}}. This GCD definition led to the modern [[abstract algebra]]ic concepts of a [[principal ideal]] (an ideal generated by a single element) and a [[principal ideal domain]] (a [[domain (ring theory)|domain]] in which every ideal is a principal ideal).

Certain problems can be solved using this result.<ref>{{Harvnb|Schroeder|2005|p=23}}</ref> For example, consider two measuring cups of volume {{math|''a''}} and {{math|''b''}}. By adding/subtracting {{math|''u''}} multiples of the first cup and {{math|''v''}} multiples of the second cup, any volume {{math|''ua'' + ''vb''}} can be measured out. These volumes are all multiples of {{math|1=''g'' = gcd(''a'', ''b'')}}.