prove that 3upon 2root3 is an irrational no.
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Can you prove that 3÷2 root 3 is an irrational number?
Assume that 3/(2*(3^.5)) is rational. Then we can write
3/(2*(3^.5) = a/b where a and b are rational.
Then 3 *b = a*2*(3^.5)
So ((3*b)/(a*2)) = (3^.5). But the left hand side is a rational expression of rational numbers, so root 3 must also be rational. But it isn’t. So the initial assumption that 3/(2*(root 3)) was rational must be false, and, therefore it must be irrational.
Note that this argument would apply to any expression that was a rational expression of rational number and a single irrational number. (The later condition is needed because the product (or ratio) of two irrational numbers can be rational. (E.g., (root 3)*(root 3) is clearly rational, provided that you’ve defined “root 3” to be the square root of 3.
Assume that 3/(2*(3^.5)) is rational. Then we can write
3/(2*(3^.5) = a/b where a and b are rational.
Then 3 *b = a*2*(3^.5)
So ((3*b)/(a*2)) = (3^.5). But the left hand side is a rational expression of rational numbers, so root 3 must also be rational. But it isn’t. So the initial assumption that 3/(2*(root 3)) was rational must be false, and, therefore it must be irrational.
Note that this argument would apply to any expression that was a rational expression of rational number and a single irrational number. (The later condition is needed because the product (or ratio) of two irrational numbers can be rational. (E.g., (root 3)*(root 3) is clearly rational, provided that you’ve defined “root 3” to be the square root of 3.
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