prove that there is a positive number x such that x^3 = 5 Solution consider the

Question

prove that there is a positive number x such that x^3 = 5

Explanation / Answer

consider the set S = {x| x^3 < 5} now S is bounded from above as x^3 is a strictly increasing function and 5^3 > 5 Hence S has a least upper bound say k now lets evaluate k^3 now if k^3 5 Lets consider number (k-a)^3 = k^3 - 3*k^2*a + 3*k*a^2-a^3 > k^3 - 3*k^2*a since k>a now k^3 - 3*k^2*a we can choose a which is very small then we can fit k^3 > k^3 - 3*k^2*a > 5 by choosing small enough a we fit k^3 - 3*k^2*a in between k^3 and 5 Hence we can find a smaller upper bound [k-a] hence its not the least upper bound. contradiction Hence we must have k^3=5 hence proved. message me if you have any doubts

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