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My calc 3 professor showed us this the other day and I thought I would try to explain it to you. When a number is irrational it means that it cannot be represented as a fraction of two integers nor can it be represented as a terminating or repeating decimal number. Pi is irrational. It is a decimal number that will go on forever and is non repeating. So to prove that [ch8730]2 is irrational you can justify an incorrect condition and then contradict it.
So say [ch8730]2 is rational.
Then [ch8730]2=p/q The fraction p/q is irreducible. (This is important you will see why at the end).
p=[ch8730]2q
p[sup]2[/sup]=2q[sup]2[/sup] 2q[sup]2[/sup] must be an even number. Therefor p[sup]2[/sup] must be even and p must also be even. Squaring an odd number yields a odd number, it's the same with even numbers too.
So since p is even we can say p=2k k must be an integer.
By substituting we have (2k)[sup]2[/sup]=2q[sup]2[/sup]
4k[sup]2[/sup]=2q[sup]2[/sup]
2k[sup]2[/sup]=q[sup]2[/sup] By the logic explained above we can conclude that q must also be even.
so [ch8730]2=p/q but if p and q are even then the fraction p/q is reducible by at least 2.
By contradicting the initial condition that p and q are not coprime
[ch8730]2 must not be rational because the math showing it could be represented as a irreducible fraction contradicted itself.
Isn't math fun! ;D
-Tony
So say [ch8730]2 is rational.
Then [ch8730]2=p/q The fraction p/q is irreducible. (This is important you will see why at the end).
p=[ch8730]2q
p[sup]2[/sup]=2q[sup]2[/sup] 2q[sup]2[/sup] must be an even number. Therefor p[sup]2[/sup] must be even and p must also be even. Squaring an odd number yields a odd number, it's the same with even numbers too.
So since p is even we can say p=2k k must be an integer.
By substituting we have (2k)[sup]2[/sup]=2q[sup]2[/sup]
4k[sup]2[/sup]=2q[sup]2[/sup]
2k[sup]2[/sup]=q[sup]2[/sup] By the logic explained above we can conclude that q must also be even.
so [ch8730]2=p/q but if p and q are even then the fraction p/q is reducible by at least 2.
By contradicting the initial condition that p and q are not coprime
[ch8730]2 must not be rational because the math showing it could be represented as a irreducible fraction contradicted itself.
Isn't math fun! ;D
-Tony