MHB How to show uniqueness in this statement for integers

Click For Summary
The statement claims there exists a unique integer n such that n^2 + 2 = 3. The proof shows that n can be either -1 or +1, indicating two valid integer solutions. This demonstrates that the solution is not unique, as both integers satisfy the equation. The discussion emphasizes that the uniqueness fails when considering all integers, including negatives. Therefore, the statement is disproven by the existence of multiple solutions.
cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
Dear Everyone,

Directions: Decide whether the statement is a theorem. If it is a theorem, prove it. if not, give a counterexample.

There exists a unique integer n such that $$n^2+2=3$$.

Proof:
Let n be the integer.

$$n^2+2=3$$
$$n^2=1$$
$$n=\pm1$$

How show this is unique or not? Please explain why if not.
 
Mathematics news on Phys.org
Cbarker1 said:
Dear Everyone,

Directions: Decide whether the statement is a theorem. If it is a theorem, prove it. if not, give a counterexample.

There exists a unique integer n such that $$n^2+2=3$$.

Proof:
Let n be the integer.

$$n^2+2=3$$
$$n^2=1$$
$$n=\pm1$$

How show this is unique or not? Please explain why if not.

Hi Cbarker1,

You found 2 integer solutions that indeed satisfy the equation.
Doesn't that mean that the solution is not unique?
Our counter example is the fact that both n=-1 and n=+1 are solutions.
It would be different if we were only looking at natural numbers, excluding negative numbers, but 'just' integers can be negative.
 
I like Serena said:
Hi Cbarker1,

You found 2 integer solutions that indeed satisfy the equation.
Doesn't that mean that the solution is not unique?
Our counter example is the fact that both n=-1 and n=+1 are solutions.
It would be different if we were only looking at natural numbers, excluding negative numbers, but 'just' integers can be negative.

Yes, since I found two solutions to satisfy the equation. therefore, n is not unique.
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 35 ·
2
Replies
35
Views
5K
Replies
1
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
7K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
27
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K