At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
So let's assume that the square root of 6 is rational. By definition, that means there are two integers a and b with no common divisors where: a/b = square root of 6. But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even.
Let's assume that it's rational: [tex]\sqrt6=\frac{p}q,p\wedge q=1[/tex]
Then [tex]p^2=6q^2[/tex] hence [tex]p[/tex] is even, which can be written [tex]p=2p'[/tex]
Hence [tex]4p'^2=6q^2[/tex] thus [tex]2p'^2=3q^2[/tex]
Thus p,q are both even, which is absurd.
Hence sqrt(6) is irrational
Then [tex]p^2=6q^2[/tex] hence [tex]p[/tex] is even, which can be written [tex]p=2p'[/tex]
Hence [tex]4p'^2=6q^2[/tex] thus [tex]2p'^2=3q^2[/tex]
Thus p,q are both even, which is absurd.
Hence sqrt(6) is irrational
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.