- #1
Darren73
- 8
- 0
Hi,
I saw a derivation in a book and I don't see the logical connection. Suppose [itex]1.\ \ a=b \text{ and } 2.\ \ x=y[/itex]
Then [itex]3.\ \ a-b= \lambda (x-y) [/itex] makes sense to me since [itex] 0=\text{Anything}⋅0[/itex]
However they said "similarly" [itex]4.\ \ a+b= \mu (x+y) [/itex], and this I don't understand. To me [itex] a+b= 2a \text{ or }2b[/itex] and [itex] x+y=2x\text{ or }2y[/itex], so unless there was a defined linear relationship such as [itex]a \propto x \text{ or }y[/itex], OR [itex]b \propto x \text{ or } y[/itex], then I don't see how they could propose equation 4. to be necessarily true.
I may have a blind spot, if so please help.
I saw a derivation in a book and I don't see the logical connection. Suppose [itex]1.\ \ a=b \text{ and } 2.\ \ x=y[/itex]
Then [itex]3.\ \ a-b= \lambda (x-y) [/itex] makes sense to me since [itex] 0=\text{Anything}⋅0[/itex]
However they said "similarly" [itex]4.\ \ a+b= \mu (x+y) [/itex], and this I don't understand. To me [itex] a+b= 2a \text{ or }2b[/itex] and [itex] x+y=2x\text{ or }2y[/itex], so unless there was a defined linear relationship such as [itex]a \propto x \text{ or }y[/itex], OR [itex]b \propto x \text{ or } y[/itex], then I don't see how they could propose equation 4. to be necessarily true.
I may have a blind spot, if so please help.