Show that square root of 3 is an irrational number

In summary, the statement that ##\sqrt{3}## is irrational is true. Its negation, that ##\sqrt{3}## is rational, is false. This is proven using the fundamental theorem of arithmetic, which states that every integer is uniquely represented as a product of prime numbers. The contradiction arises when attempting to represent ##a^2## and ##b^2##, the squares of ##a## and ##b##, as products of prime numbers, as the right hand side of the equation contains an odd number of ##3##s. Therefore, the original statement holds true.
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Homework Statement
Show that ##\sqrt{3}## is irrational
Relevant Equations
The fundamental theorem of arithmetic
##\sqrt{3}## is irrational. The negation of the statement is that ##\sqrt{3}## is rational.

##\sqrt{3}## is rational if there exist nonzero integers ##a## and ##b## such that ##\frac{a}{b}=\sqrt 3##. The fundamental theorem of arithmetic states that every integer is representable uniquely as a product of prime numbers, up to the order of the factors. So ##a## and ##b## are products of prime numbers.
$$\frac{a}{b}=\sqrt{3} \Rightarrow a^2=3b^2$$
Squaring ##a## and ##b## leads to the prime factors ## a^2## and ##b^2## existing in pairs. But, the right hand side is multiplied by ##3##, which means there is an odd number of ##3## in the right hand side. This leads to a contradiction because there is an even number of ##3##s on the left hand side.

So the negated statement is false, and hence the original statement is true.
 
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Looks good. :oldwink:
 
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Related to Show that square root of 3 is an irrational number

What is an irrational number?

An irrational number is a real number that cannot be expressed as a ratio of two integers. This means that it cannot be written as a fraction with a finite number of digits in the numerator and denominator.

How do you prove that the square root of 3 is irrational?

The proof for the irrationality of the square root of 3 is based on the fundamental theorem of arithmetic, which states that every positive integer can be expressed as a unique product of prime numbers. By assuming that the square root of 3 is rational, we can use this theorem to show that it leads to a contradiction, thus proving that it is irrational.

Can you provide an example of a proof for the irrationality of the square root of 3?

One example of a proof for the irrationality of the square root of 3 is the proof by contradiction. It starts by assuming that the square root of 3 is rational, and then proceeds to show that this leads to a contradiction. This contradiction proves that the initial assumption was false, and therefore the square root of 3 must be irrational.

Why is it important to prove that the square root of 3 is irrational?

Proving that the square root of 3 is irrational is important because it helps us understand the properties of real numbers and their relationships. It also has practical applications in various fields of mathematics, such as number theory, geometry, and algebra.

Are there any other methods for proving the irrationality of the square root of 3?

Yes, there are other methods for proving the irrationality of the square root of 3, such as the proof by infinite descent, proof by continued fractions, and proof by the rational root theorem. Each method has its own unique approach, but they all lead to the same conclusion that the square root of 3 is irrational.

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