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Prove with induction that every finite nonempty set of real numbers has a largest element.

My idea (please fix my notation where it is wrong)

Let $A=\left\{a_i\in \mathbb{R}:i\in N)\right\}$. If $i = 1, $ then we conclude that $a_1$ is the largest element. Hence we assume that A has a largest number for all N.

Now let

$$A=({a_1,a_2,...,a_k,a_{k+1}})=(a,a_{k+1}).$$

If $a_{k+1}$ is the largest element we are done, if a is the largest one then by our hypothesis we conclude that for $a_i$ for some i is the largest element. This prove also relies on the elements being well ordered.

Would this be a sufficient proof and how should it be written?

Klangen
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ALEXANDER
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    It's almost perfect: When you specify $A=({a_1,a_2,...,a_k,a_{k+1}})=(a,a_{k+1})$. What's your $a$ here ? You should first say that, with the induction hypothesis, there is a $a$ that is the maximum in $(a_1,\ldots, a_k)$ and then decide whether $a_{k+1}$ or $a$ is bigger – H. Potter May 06 '16 at 05:41
  • That would be all the $a_1$,$a_2$.... all the way up to $a_k$ – ALEXANDER May 06 '16 at 05:42
  • All right, got you. Thanks – ALEXANDER May 06 '16 at 05:43
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    I posted my comment too early, so I had to edit it. But here, in your proof, when you're saying "if $a_{k+1}$ is the largest, we're done", you're using what you want to prove. The proof is clearer if you first find the largest number among the $k$ first entries, and then compare it with $a_{k+1}$ – H. Potter May 06 '16 at 05:46
  • Okey, I guess that's what I meant, but I see what you are getting at. Good that you point that out! Thank you – ALEXANDER May 06 '16 at 05:50
  • It doesn't really depend on the elements being well-ordered.It's a proof by induction, starting with the validity in the cases of a set with 1 or 2 members. – DanielWainfleet May 06 '16 at 07:08

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