# How did we get $W_g=W_b$ from $\dfrac{U'(W_g)}{U'(W_b)}=1$?

My question is from Nicholson-Snyder's text , E-book here.

My question is here, from page 217 of the book. (I can't post image as my reputation is not enough.)

How did we get $W_g=W_b$ from $\dfrac{U'(W_g)}{U'(W_b)}=1$ ?

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@ honso : maybe from the injectivity of $U'$ ? –  TheBridge Dec 22 '13 at 18:31

$U$ is usually strictly concave. So $U'$ is strictly decreasing and is therefore injective.
@Nic: Yes, you are right but it does not contradict what I said. I said $U'$ (not $U$) is strictly decreasing. –  Hans Dec 23 '13 at 14:55