If $\omega + 1 = \omega$, find $\omega$ ($\omega \not= - \infty$ or $\infty$)

Question

If $\omega + 1 = \omega$, find $\omega$ ($\omega \not= - \infty$ or $\infty$). It does not have to be a real number.

My teacher gave us this question just to play around with, and my first guess was there was no value because for any value of $\omega$

$$ \omega + 1 = \omega \\ \omega - \omega + 1 = 0 \\ 1 = 0 $$

and $1 \not= 0$ (I think).

Then I tried to assume that $\omega = \infty$, but unfortunately I can't use that.

So then I tried another way..

$$ \omega +1 = \omega \\ \omega -\omega + 1 =0 \\ \omega(1-1) = -1 \\ \frac{\omega(1-1)}{-1} = 1 \\ -(1-1) = \frac{1}{\omega} \\ \therefore 0 = \frac{1}{\omega} \text{ where 1 has to equal 0..} $$ But this is so wrong, $1$ does not equal $0$. So can anyone find a value (if one exists) for $\omega$ that would satisfy the equation above? Thanks in advance!

Answer 1

I guess the question must be find $\omega$ if $$(\omega+1)^n=\omega^n$$ Because he mentioned that $\omega$ don't have to be real $$(\omega+1)^n=\omega^n$$ $$\omega^n(1+\frac{1}{\omega})^n=\omega^n$$Since $\omega\neq 0$ $$\left(1+\frac{1}{\omega}\right)^n=1$$ $$(z)^n=1$$ $$z=1^{1/n}$$ $$z=\left[\cos(2k\pi)+i\sin(2k\pi)\right]^{1/n}$$ $$z=\cos\left(\frac{2k\pi}{n}\right)+i\sin\left(\frac{2k\pi}{n}\right)$$ $$z=e^{2ki\pi/n}$$ since $z=1+\frac{1}{\omega}$ $$\omega=\frac{1}{z-1}$$

$$\omega=\frac{1}{e^{2ki\pi/n}-1}$$

Answer 2

Most likely your teacher is thinking of $\omega=-\infty$.

Answer 3

If it was $1+\omega=\omega$, then the ordinal $\omega$ could be the answer. Are you sure you did not reverse the sum ?

Answer 4

I do not certain that this 'answer' presents the answer which your teacher wants, since 'infinite cardinals' are some type of infinity.

In the case of finite sets, we can count the number of elements of the set and we can compare the 'size' of two sets. For example, $A=\{25,39,93,1341,9999\}$ and $B=\{h,m,n,r,x\}$ have same number of elements, and in fact, they have 5 elements.

We can suggest the question naturally - we can compare the size of finite sets, but how to compare the size of infinite sets? We cannot 'count' the number of elements of infinite sets. However, if $A$ and $B$ have same 'size', we can correspond the elements of $A$ to the elements of $B$. For example, in previous example we can correspond the elements of $A$ to $B$ as follows: $$25\mapsto n,\, 39\mapsto h,\, 93\mapsto m,\, 1341\mapsto r\text{ and } 9999\mapsto x$$

It is same for the case of infinite sets - that is, if there is a 1-1 correspondence between $A$ and $B$, we can consider that they have same size. Mathematicians (exactly, set theorists) define a variety of kinds of infinite sizes. For example, we can think the size of the set of natural numbers. We call it aleph-zero and denotes $\aleph_0$, and it is the least infinite cardinal.

In the case of finite numbers, we can think the sum of sizes. For example, we can think $4+4$ and we know that it is equal to 8. Like the finite numbers, we can also think and define the sum of infinite cardinals. How to define it? At first, we consider the finite case. If $A$ and $B$ are disjoint, (that is every element of $A$ is not an element of $B$) then $|A\cup B|=|A|+|B|$. We can define the addition of infinite sizes similarly. Even if $A$ and $B$ are not disjoint, we can make it disjoint. (For example, we can consider the set $\{(a,0):a\in A\}$ and $\{(b,1):b\in B\}$ instead of $A$ and $B$.) From above discussion, we can define $\aleph_0+1$ and especially, we can get $\aleph_0+1=\aleph_0$.