- #1

docnet

- 587

- 242

- 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.

##\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.