0

$(A):$For any $k\in \Bbb R,$for any $n\in \Bbb N$ \ {$0$} ,there exists some $z\in \Bbb C$ such that $k^2+n\lt |z|$.

$(B):$For any $n\in \Bbb N$ \ {$0$},there exists some $z\in \Bbb C$ such that for any $k\in \Bbb R,k^2+n\lt |z| $.

The first part of the question is to write down the negations of the above statements, so I have written down

$(\sim A):$There exists some $k\in \Bbb R$ and $n\in \Bbb N$ \ {$0$} such that for any $z\in \Bbb C$, $k^2+n\geq |z|$.

$(\sim B):$There exists some $n\in \Bbb N$ \ {$0$} such that for any $z\in \Bbb C$,there exists some $k\in \Bbb R$ such that $k^2+n\geq |z| $.

I am not sure if the above is correct. The second part of the question is to justify the statement A,B is true or not. However, I do not know how to prove them because the two statements are so similar, I am confused what is the difference between A and B. (How can I make use of negations to solve the problem?)

Can I just casually pick any value of $k,n,z$ to finish the proof?

Thanks.

sunny
  • 133

1 Answers1

1

Your analysis is good, as far as you took it.

Part A is true, and Part B is false.

For Part A:

For any $k \in \Bbb{R},$ and positive integer $N$, you have that the sum $k^2 + N$ is a finite positive number.

For any finite positive number $M$, consider, in the Complex plane, the circle of radius $M+1$ (for example), centered at $[0 + i(0)]$. Then, any element $z$ on this circle will be such that $|z| = M+1.$

Even more simple is to accept that $\Bbb{R} \subset \Bbb{C}.$ Therefore, given any fixed sum $k^2 + N$, the Complex number $z = [(k^2 + N + 1) + i(0)]$ will be such that $|z| > k^2 + N.$

For Part B:

The easiest way to demonstrate that Part B is false is to attempt to use the analysis above, which applies to Part A, to also prove Part B.

If you do this, you will see that the critical distinction between Parts A and B is that in Part A, you choose $N$ and $k$ first, and then go looking for some satisfying complex number $z$.

In part B, the analysis used for Part A is therefore invalid. This is because you are given $N$. Then, you have to find some value $|z|$ that will be satisfactory, no matter what value of $k$ is chosen, after $z$ is selected.

The following analysis demonstrates that part B is false:

Given any positive $N$, and given any $z$, you must have that $|z| - N$ is some fixed finite number. This means that regardless of what value of $z$ you propose as a candidate satisfying value, you can simply choose $k \in \Bbb{R}$ such that $k^2 > |z| - N.$

user2661923
  • 35,619
  • 3
  • 17
  • 39