Find all real numbers x satisfyingfloor(x+5) = floor(3x)HINT: Every real number x can be written as n + E where n is an integer and E is a real number such that 0 ≤ E ≤ 1
Hello, It bears repeating: you should have applied the well-known following formulas: Hello, Please and Thanks. Anyway... = = = = = = = = = = = = = = = = = = Let x be written as n+ε where n is an integer and ε is a real number such that 0≤ε<1. Then: ???x + 5 = n + ε + 5 = n + 5 + ε ???floor(x + 5) = n + 5 ???3x = 3(n + ε) = 3n + 3ε ???floor(3x) = 3n + floor(3ε) → If 0≤ε<? then ???0 ≤ 3ε < 1 ???floor(3ε) = 0 ???n + 5 = 3n ???2n = 5 ???n = 5/2 BUT n must be an integer thus this solution is invalid. → If ?≤ε<? then ???1 ≤ 3ε < 2 ???floor(3ε) = 1 ???n + 5 = 3n + 1 ???2n = 4 ???n = 2 THUS all reals of the interval [ (2+?); (2+?) ) are solutions. → If ?≤ε<1 then ???2 ≤ 3ε < 3 ???floor(3ε) = 2 ???n + 5 = 3n + 2 ???2n = 3 ???n = 3/2 BUT n must be an integer thus this solution is invalid. CONCLUSION: All real numbers x satisfying floor(x+5)=floor(3x) are the reals of interval [7/3; 8/3). Regards, Dragon.Jade :-)
