Answer to Question #102134 in Calculus for Kampamba

Question #102134
Prove that root of 2 plus root of 3 is irrational
1
Expert's answer
2020-01-31T10:52:46-0500

Solution. We prove that the number is irrational by contradiction. Let "\\sqrt{2}+\\sqrt{3}" be rational.

Therefore, there exist coprime integers "a" and "b":


"\\sqrt {2}+\\sqrt{3} = \\frac{a}{b}"

Raised to a power 2 both sides of the equation one gets


"2+2\\sqrt6+3=\\frac{a^2}{b^2}"

"\\sqrt6=\\frac{a^2}{2b^2}-\\frac{5}{2}"

"\\sqrt6=\\frac{a^2-5b^2}{2b^2}"

"a^2-5b^2" and "2b^2" are integers because "a" and "b" are integers. Hence, the number "\\sqrt{6}" is rational. Thus, we can write


"\\sqrt6=\\frac{p}{q}"

where "p,q" are coprime integers. From the last equation it follows that


"6q^2=p^2"

The left side of the equation is divisible by 2 therefore the right side is divisible by 2, hence "p" is an even number. As a result,


"p=2k"

where "k" is integer. Therefore,


"6q^2=4k^2 \\implies 3q^2=2k^2"

The right side of the last equation is divisible by 2, hence the left side is divisible by 2. As a result,


"q=2n,"

where "n" is integer. Therefore, one gets "p, q" are not coprime integers. A contradiction occurred. Hence, "\\sqrt{6}" is irrational and

the assumption that 'the number "\\sqrt{2}+\\sqrt{3}" is rational' is false. We proved that the number "\\sqrt{2}+\\sqrt{3}" is irrational by contradiction.







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