Proof questions that I wanna make sure I got right! (1 Viewer)

[Run hard@thehsc]

 

[yanujw]

8. Suppose the two factors are a, b > root(n)
ab > n, which is a contradiction since ab = n as we define them as a pair of factors to n.

9.
a) n^2 = 4m - 2
n^2 = 2(2m-1)
The RHS is even, implying the LHS is even. For n^2 to be even, n is even.

b) Let n=2p.
4p^2 + 4 = 2
2p^2 +2 = 1
2(p^2 +1) = 1
The LHS is even, while 1 is odd. Hence there is a contradiction and no n satisfies the condition.

 

[Run hard@thehsc]

Got it! Thank you...

 

[GhoulGhost]

8. Suppose the two factors are a, b > root(n)
ab > n, which is a contradiction since ab = n as we define them as a pair of factors to n.

9.
a) n^2 = 4m - 2
n^2 = 2(2m-1)
The RHS is even, implying the LHS is even. For n^2 to be even, n is even.

b) Let n=2p.
4p^2 + 4 = 2
2p^2 +2 = 1
2(p^2 +1) = 1
The LHS is even, while 1 is odd. Hence there is a contradiction and no n satisfies the condition.

Just a quick note, "For n^2 to be even, n is even." also works for testing whether n is divisible by other integers if the integer itself has square-free factors.