I was taking a look at some of the practice question from the first chapter of Spivak, and I am wanting to just verify if I am on the right track with things. I do not have solutions, so I am just trying to see if there is anything I am missing, or even if I am completely off and need to approach them differently.

For an example of one of the basic problems, ( to show the methods I use for thsese type of questions) For another example, one questions asks

$\mathbf{Q:}$ if $$a \lt b$$ and $$c \lt d$$ then prove that $$a+c \lt b+d$$

Then should I just use what is given by the trichonomy law and such, ( P denotes positive numbers)

ie, I would say

If $a \lt b$ , then $b-a \in P$

and further $(b-c)+(c-a) \in P$

ie $a+c \lt b+c$

and if $c \lt d$, then $d-c \in P$

and by the same approach $c+b \lt d+b$

Which seemingly would show $a+c \lt b+d$

However, I am unsure if this is the correct approach or if it is containing enough rigour to be valid. Any insight/tips/etc? Thanks

  • 1
    $\begingroup$ This is completely rigorous. I think you may have made a typo. Your last line should be $a+c < b+d$, or perhaps $a+c<b+c<b+d$. $\endgroup$ – Alex G. Jun 23 '15 at 20:16

You can also go a more direct route. Suppose $a<b$ and $c<d$, then $b-a\in P$ and $d-c\in P$.

By closure under addition (P11), $(b-a)+(d-c)\in P$, and so $$(b-a)+(d-c)=b+d-a-c=b+d-(a+c)$$

thus, $a+c<b+d$.


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