Prove using only integers and inequalities that there is no natural number whose

Question

Prove using only integers and inequalities that there is no natural number whose square is 10. (Producing a decimal approximation of square 10, or arguing that squareroot 10 is between 2 and 3 is not proof. Just because one solution of x^2 = 10 is not an integer doesn't mean that there couldn't be some other that is an integer. Do not rework the method that is given in many textbooks for showing that squareroot s of non-perfect squares are irrational. Do not use number-theoretic concepts such as divisibility or relative primarily. Your solution must contain no references to squareroot s or quotients, or use algebraic facts concerning the solutions of quadratic equations.)

Explanation / Answer

10<16

Hence,sqrt{10}<4

9<10

Hence, 3<sqrt{10}<4

But there is not natural number between 3 and 4

Hence there is no natural number who square is 10

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