- #1
SOHAWONG
- 16
- 0
when X is even number,it's easy to prove
but how about the condition which X is odd number?
I have no idea of this
but how about the condition which X is odd number?
I have no idea of this
no,i may add despite 1,4,9,16,25...etcHurkyl said:[itex]\sqrt{4}[/itex] is irrational?
yes, but how to prove?Char. Limit said:So in other words...
[tex]\sqrt{x}[/tex] is irrational iff x=/=n^2 for n belonging to the integer set.
Tinyboss said:Fundamental theorem of arithmetic. Assume p^2/q^2=x with gcd(p,q)=1, and see what has to divide what.
An irrational number is a real number that cannot be expressed as a ratio of two integers. It is a non-repeating, non-terminating decimal number.
The most commonly used method to prove that √X is irrational is by contradiction. Assume that √X is rational and can be expressed as a ratio of two integers, then use algebraic manipulation to reach a contradiction.
The proof by contradiction method assumes the opposite of what is to be proven and then shows that this assumption leads to a contradiction, thus proving the original statement to be true.
Assume that √2 is rational and can be expressed as a/b, where a and b are integers with no common factors. Then, √2 = a/b can be rewritten as 2 = a^2/b^2, which means 2b^2 = a^2. This implies that a^2 is even, and therefore a must be even. Let a = 2c, where c is another integer. Substituting this into the equation 2b^2 = a^2, we get 2b^2 = (2c)^2 = 4c^2. This means that b^2 = 2c^2, and similarly, b must also be even. However, this contradicts our initial assumption that a and b have no common factors. Therefore, √2 cannot be expressed as a ratio of two integers, and it is irrational.
Yes, there are other methods such as the continued fraction method and the Euclidean algorithm. However, the proof by contradiction method is the most commonly used and relatively straightforward.