Linear Algebra Nonparallel vector Proof

SMG75
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Homework Statement


Suppose x, y \in\Re^{n} are nonparallel vectors.
a) Prove that if sx+ty=0, then s=t=0
b) Prove that if ax+by=cx+dy, then a=c and b=d

Homework Equations


The Attempt at a Solution


I'm very new to proof-based math, so I'm just trying to get my feet wet here. I realize this is a very simple problem, but I was hoping someone could walk me through it.

Assuming s\neq0, then sx=-ty. Thus, x=(-t/s)y, which goes against the definition of nonparallel vectors. (Two vectors are parallel if they can be written with scalar constant. i.e. x=cy or y=cx). So, as a result, we know s\neq0 must not be possible. Using similar logic, t=0 cannot be possible, either. Therefore, s=t=0.

I realize this is incredibly rough; I'm just looking to see if my thinking is on the right track.

Thanks!
 
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SMG75 said:

Homework Statement


Suppose x, y \in\Re^{n} are nonparallel vectors.
a) Prove that if sx+ty=0, then s=t=0
b) Prove that if ax+by=cx+dy, then a=c and b=d


Homework Equations





The Attempt at a Solution


I'm very new to proof-based math, so I'm just trying to get my feet wet here. I realize this is a very simple problem, but I was hoping someone could walk me through it.

Assuming s\neq0, then sx=-ty. Thus, x=(-t/s)y, which goes against the definition of nonparallel vectors. (Two vectors are parallel if they can be written with scalar constant. i.e. x=cy or y=cx). So, as a result, we know s\neq0 must not be possible. Using similar logic, t=0 cannot be possible, either. Therefore, s=t=0.

I realize this is incredibly rough; I'm just looking to see if my thinking is on the right track.
You're on the right track. What you're doing is a proof by contradiction, which is where you prove that P == Q by assuming that P is true and that Q is false. If you arrive at a contradiction, then you have indirectly proved that P ==> Q.

For this problem you should assume that both s and t are nonzero. That way you can show that x = (-t/s) y and that y = (-s/t) x. Since, by assumption x and y are nonparallel, you have arrived at a contradiction.
 
For the second part, group the x terms together and the y terms together.
 
Notice also that for the second problem, you may find it helpful to use the result of the previous problem. The proof will follow quite naturally from this. Good work so far!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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