# Problem with base case for transfinite induction

I need to prove this using transfinite induction

Let $\alpha, \beta , \gamma$ ordinals

If $\beta <\gamma$ then $\alpha + \beta < \alpha + \gamma$

I am trying to prove the statement by transfinite induction on $\gamma$

for base case $\gamma =0$, I assume that $\beta <0$, but as there is no $\beta<0$ then $\alpha +\beta < \alpha +0$ holds for all $\beta$ ? is that correct ?

• but what if I try to prove that $\alpha + \beta > \alpha + \gamma$ (something that is false), would no be the same argument correct for the base case $\gamma =0$? Mar 21, 2015 at 1:47
• I don't understand what you're saying,but here are some possibly relevant observations. It obviously is false that $\forall \alpha, \beta,\gamma(\alpha+\gamma<\alpha+\beta)$. It is also false that $\forall \alpha,\beta(\alpha+0<\alpha+\beta)$. But it is not false that $\forall \alpha,\beta(\beta <0\to \alpha+0<\alpha+\beta)$. Mar 21, 2015 at 1:52
If you take the base case to be $\gamma=0$ (instead of the more common $\beta=0$), you basically shift the problem to the induction step.
To see this, suppose you have shown the inequality for $\gamma_0$, and you want to show it for $\gamma_0+1$. Now $\beta<\gamma_0+1$. You want to use the induction hypothesis, but now we only know $\beta\le\gamma_0$, which means either $\beta<\gamma_0$ or $\beta=\gamma_0$. The first case is handled by the induciton hypothesis, but the second is not, and you have to check it manually. Now $\gamma=\gamma_0+1$, so the inequality you want to check is $\alpha+\beta<\alpha+\gamma_0+1$. This is true because $\beta=\gamma_0$. However, if you reverse the inequality to be shown, then it is false in this special case that is not reduced to the induction hypothesis.