Square Root of 3Date: 05/18/99 at 02:35:50 From: sheila Subject: Pure Maths What is the appropriate way to prove that the square root of 3 is an irrational number? Date: 05/18/99 at 13:28:50 From: Doctor Nick Subject: Re: Pure Maths Hi Sheila - Here's one way to prove this. Suppose that the square root of 3 is r/s, where r and s are relatively prime positive integers (if the square root of 3 is rational, then we can do this). Squaring both sides of the equation yields 3 = (r^2)/(s^2). That is, 3 s^2 = r^2. This shows that r^2 must be divisible by 3, from which it follows that r is divisible by 3, and r^2 is divisible by 9. This implies that s^2 must be divisible by 3, and so s must be divisible by 3. Hence, r and s are divisible by 3. This is a contradiction to our assumption that r and s are relatively prime. Thus, there must not exist integers r and s so that the square root of 3 is r/s. In other words, the square root of three is irrational. You can use an almost identical proof to show that the square root of any prime is irrational. With a little modification, you can use it to show that the square root of an integer is either an integer or an irrational number. Have fun, - Doctor Nick, The Math Forum http://mathforum.org/dr.math/ |
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