- #1
docnet
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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.
##\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.
