Week 8
SundayThere will be two midterms for me next week: CSC165 and CSC148. And I was too busy to wrote any blog for the past two weeks due to the amount of assignments I had. But in this blog, I want to take chance to talk about Assignment 2 which I had last week.
Overall, the assignment was not that hard and it took me about 2.5 hours to finished it and checked for a second time. However, for question 5 in the assignment:
4. Prove or disprove: for all quadruples of positive real number
w, x, y, z, if w / x < y / z, then:
( w / x < ( w + y ) / ( x + z ) ) ^ ( ( w + y ) / ( x + z )
< y / z )
A: Prove/ true
assume ∀ w, x, y, z ∈ R, w > 0, x > 0, y > 0, z > 0
and w / x < y / z
=> ( w / x < ( w + y ) / ( x + z ) ) ^ ( ( w + y ) / ( x +
z ) < y / z )
assume ( w / x < ( w + y ) / ( x + z ) )
=> ( ( w / x ) * ( x ( x + z ) ) ) < ( ( w + y ) / ( x + z
) * ( x ( x + z ) )
then = ( x + 2 ) w <
x ( w + y )
= x w + z w < x w + x y
= z w < x y
= z w / x < y
= w / x < y / z
therefore, it is true
assume ( ( w + y ) / ( x + z ) < y / z )
=> ( ( w + y ) / ( x + z ) * z ( x + z ) ) < ( y / z * ( z
( x + z ) ) )
then = ( w + y ) z <
y ( x + z )
= w z + z y < y x + z y
= w z < y x
= w / x < y /z
therefore, it is true
therefore, ∀ w, x, y, z ∈ R, w > 0, x > 0, y > 0, z
> 0, w / x < y / z =>
( w / x
< ( w + y ) / ( x + z ) ) ^ ( ( w + y ) / ( x + z ) < y / z )
therefore, it is true
The above was my solution. I also checked the sample solution. But it is much different from mine: I separate the two parts and then solved each one of them. And it also seems that I have so much more steps than the sample solution because I thought just in case that I would lose marks for not having enough steps and explain my answer clearly.
Therefore, my question is that does my solution to the question look right? And I also have noticed that my writing style in the problem-solving is not exactly like the sample solutions nor notes, does it matter a whole lot and might causing me to lose marks in an exam?
Thank you very much for your time!
Roland Qiyang Liu
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