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## Homework Statement

## Homework Equations

## The Attempt at a Solution

For part (a):

a*1 + b*√2 + c*√3 = 0

assume a, b, c not all zero

a + b√2 = -c√3

a

^{2}+ 2b

^{2}+ 2ab√2 = 3c

^{2}

a

^{2}+ 2b

^{2}- 3c

^{2}= -2ab√2

(a

^{2}+ 2b

^{2}- 3c

^{2})/(-2ab) = √2

which is not possible since we take a, b, c to be rational, and √2 is irrational.

thus our assumption of a, b, c not all zero was false and we must have a=b=c=0.

For part (b), a similar argument, but easier:

{1, 1 + √5, (1 + √5)

^{2}} = {1, 1 + √5, 1 + 2√5 + 5}

a + b(1 + √5) + c(1 + 2√5 + 5) = 0

assume a, b, c not all zero

1 + b + b√5 + c + 2c√5 + 5c = 0

1 + b + c + 5c = -b√5 - 2c√5

1 + b + 6c = (-b - 2c)√5

(1 + b + 6c)/(-b - 2c) = √5

same story as before.

Now that was super easy. And the assignment says to be careful with the structure of my argument. And I hate denominators because they can't be zero. If a = b = c = 0, then we have 0/0, which is of indeterminate form, which is ok! but i'm getting the idea that my argument is flawed because of division by zero...