Answer:
let 2+3[tex]\sqrt{5}[/tex] be x
let us assume that 2+3[tex]\sqrt{5\\[/tex] is rational
thus x = a/b
so [tex]\sqrt{5}[/tex] = a-2b/3
a-2b / 3 is rational number
but given that [tex]\sqrt{5\\[/tex] is irrational
thus our assumption is wrong and x is irrational
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Answers & Comments
Answer:
let 2+3[tex]\sqrt{5}[/tex] be x
let us assume that 2+3[tex]\sqrt{5\\[/tex] is rational
thus x = a/b
so [tex]\sqrt{5}[/tex] = a-2b/3
a-2b / 3 is rational number
but given that [tex]\sqrt{5\\[/tex] is irrational
thus our assumption is wrong and x is irrational