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## Main Question or Discussion Point

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?