Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am getting confused about ordinal arithmetic
I am trying to figure out what $\omega \cdot (\omega+1)$ is.
The fact that the distributive law applies from the left side makes me think that the answer is $\omega^2 + \omega$. However applying the supremum definition of multiplication I am getting a different answer. The supremum method is giving me: $$\omega \cdot (\omega+1) = \sup \{\omega \cdot n \ | \ n \in \omega+1 \} = \omega^2 $$ Since $\omega^2 \geq \omega \cdot n, \forall n < \omega + 1$.

Which way is correct and why is the other method wrong here?

share|improve this question
6  
$\omega+1$ is not a limit. It is a successor. –  Chris Eagle Jan 16 '13 at 21:33
1  
$\omega^2+\omega$ is correct: you have a string of $\omega+1$ copies of $\omega$. @Chris has identified the error in the other approach. –  Brian M. Scott Jan 16 '13 at 21:34

2 Answers 2

You are wrong in the definition of multiplication. It holds that $\alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma$, but $(\alpha+\beta)\cdot\gamma\neq\alpha\cdot\gamma+\beta\cdot\gamma$ in some of the cases.

It follows that $\omega\cdot(\omega+1)=\omega\cdot\omega+\omega\cdot1=\omega^2+\omega$.

share|improve this answer

The second method is wrong, because the definition of multiply that you used in second method is right only when the second ordinal is a limit ordinal, so if the second ordinal is something like $\omega+1$ the definition of multiply is the distributive law :$a(b+1) = ab+a$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.