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OpenStudy (watchmath):
Argue by contradiction. Suppose a > b. What would happen if we assume that?
OpenStudy (anonymous):
that means a> b + 1/n
OpenStudy (watchmath):
why?
OpenStudy (anonymous):
no!! sorry it will be na>b that lead to a>b/n then b/n>a
OpenStudy (watchmath):
Does it lead to any contradiction?
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OpenStudy (anonymous):
yes
OpenStudy (anonymous):
no what I wrote it does not lead but I wrote like if I use the property
OpenStudy (anonymous):
are you with me?
OpenStudy (watchmath):
We want to get a contradiction to the given statement. A contradiction to that statement means that there is an integer m such that a> b + 1/m . How can we get such m ?
OpenStudy (anonymous):
How?
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OpenStudy (anonymous):
I DO NOT KNOW I want to know that please
OpenStudy (watchmath):
well we need to work backward. Suppose there is such m such that a > b+ (1/m). Then a-b > 1/m. Then m > 1/(b-a). How the archimedean property fit in this situation?
OpenStudy (watchmath):
m > 1/(a-b) I meant
OpenStudy (anonymous):
we will say m+b>a
OpenStudy (anonymous):
no!!1
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