- #1

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but how about the condition which X is odd number?

I have no idea of this

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- Thread starter SOHAWONG
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In summary, when X is an even number, it is easy to prove that \sqrt{x} is irrational. However, when X is an odd number, it becomes more complicated. The condition for \sqrt{x} to be irrational is that x is not equal to n^2 for any integer n. This can be proven using the fundamental theorem of arithmetic, which states that if p^2/q^2 = x with gcd(p,q)=1, then certain numbers must divide others. The proof for \sqrt{2} can be adapted for other numbers, but it's important to note that "even" just means "divisible by 2" and is not relevant for numbers other than 2. For more information and a

- #1

- 16

- 0

but how about the condition which X is odd number?

I have no idea of this

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- #2

Staff Emeritus

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[itex]\sqrt{4}[/itex] is irrational?

- #3

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no,i may add despite 1,4,9,16,25...etcHurkyl said:[itex]\sqrt{4}[/itex] is irrational?

- #4

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[tex]\sqrt{x}[/tex] is irrational iff x=/=n^2 for n belonging to the integer set.

- #5

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yes, but how to prove?Char. Limit said:

[tex]\sqrt{x}[/tex] is irrational iff x=/=n^2 for n belonging to the integer set.

- #6

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- #7

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Tinyboss said:

what does gcd mean?

- #8

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Look at the proof for sqrt(2), and adapt it. Remember that "even" just means "is divisible by 2", so that if you're checking a number other than 2, you won't be thinking about "even" anymore.

- #9

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http://en.wikipedia.org/wiki/Square_root

http://en.wikipedia.org/wiki/Square_root_of_2

And for an interesting history of the discovery of irrational numbers look at

http://en.wikipedia.org/wiki/Irrational_number

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.

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