Proving the Existence of One Positive Integer m and n for x in Real Numbers

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The discussion centers on proving the existence of unique positive integers m and n for any real number x, where m satisfies m ≤ x < m + 1 and n satisfies n < x ≤ n + 1. Participants suggest defining two sets, X and Y, where X contains integers less than x and Y contains integers less than or equal to x. The supremum of these sets, m and n respectively, is proposed as the solution to demonstrate the required properties. The conversation emphasizes the importance of proving that m belongs to set X and n belongs to set Y. This approach provides a structured method to establish the existence of the integers in question.
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x is a real number, could you tell me how to prove that there will always be one and only one positive integer m such that m=< x <m+1 and one and only one positive integer n such that n< x =< n+1 ?
Thank you...
 
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Looks like homework, read the rules on homework above (we'll help you, but we won't do it for you).
 
Oh, Poet jcsd, I didn't know that.
But how will Poet help me ?
 
YourLooks said:
x is a real number, could you tell me how to prove that there will always be one and only one positive integer m such that m=< x <m+1 and one and only one positive integer n such that n< x =< n+1 ?
Thank you...
Suppose there are 2 sets X and Y, X={k belongs to Z : k<x}, Y={k belongs to Z : k<=x}, which means X and Y are not set zero and there should be a sup. Now you only need to let m=supX and n=supY then try to prove m belongs to X and n belongs to Y. Thats all.
 
Thank you for helping Yourlooks, you are really new to this forum :wink:
 
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