- #1
W3bbo
- 31
- 0
Starting with simple fractions, it's known that:
[tex]{{{a \over b}} \over {{c \over d}}} = {{ad} \over {bc}}[/tex]
So when b == d:
[tex]{{{a \over b}} \over {{c \over b}}} = {a \over c}[/tex]
But what if in the case of:
[tex]{{{{1 + \sqrt 2 } \over {\sqrt 2 }}} \over {{{1 - \sqrt 2 } \over {\sqrt 2 }}}}[/tex]
Following the above (for abc), it should result in:
[tex]{{1 + \sqrt 2 } \over {1 - \sqrt 2 }}[/tex]
But if you follow the abcd version:
[tex]{{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}[/tex]
...which is not equal to the first expression.
Why?
[tex]{{{a \over b}} \over {{c \over d}}} = {{ad} \over {bc}}[/tex]
So when b == d:
[tex]{{{a \over b}} \over {{c \over b}}} = {a \over c}[/tex]
But what if in the case of:
[tex]{{{{1 + \sqrt 2 } \over {\sqrt 2 }}} \over {{{1 - \sqrt 2 } \over {\sqrt 2 }}}}[/tex]
Following the above (for abc), it should result in:
[tex]{{1 + \sqrt 2 } \over {1 - \sqrt 2 }}[/tex]
But if you follow the abcd version:
[tex]{{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}[/tex]
...which is not equal to the first expression.
Why?