Proof by induction on the number of horses: Basis Step. There is only one horse. Then clearly all horses have the same color. Induction Hypothesis. In any group of up to n horses, all horses have the same color. Induction Step. Consider a group of n 1 horses. Discard one horse; by the induction hypothesis, all the remaining horses have the same color. Now put that horse back and discard another; again all the remaining horses have the same color. So all the horses have the same color as the ones that were not discarded either time, and so they all have the same color.

Claiming mathematical induction, of the statement: "all horses are the same color", the theorem is a counterfeit paradox sustained by mistaken  demonstrations.

Step-by-step explanation:

”that is a horse of a different  color” was a familiar expression in the middle of the last century, meaning that something is quite different from normal or common expectation, but George Polya, a great mathematician provided proof that there is no horse of a different color:

Theorem: "All horses are the same color"

Proof (by induction on the number of horses):

- Base Case: P(1) is undoubtedly true, as having only one horse, then all horses have the same color.

- Inductive Hypothesis: Assume P(n), which is the statement that n horses all have the same color.

- Inductive Step: Given a set of n+1 horses {h1,h2,...,hn+1}, we can eliminate the last horse in the serie  and use the inductive hypothesis onlky to the first n horses {h1,...,hn}, deducing that they all have  the same color. The same way, the conclusion may be that the last n horses {h2,...,hn+1} all have the same  color. But the “middle” horses {h2,...,hn} (i.e., all but the first and the last) belong to both of  these series, so they have the same color as horse h1 and horse hn+1. It follows, therefore, that all n+1  horses have the same color. Therefore, using the principle of induction, all horses have the same color.

It is clear that, it is not true that all horses are of the same color, so where is the mistake in our induction  proof? It is tempting to blame the induction hypothesis. But even though the induction hypothesis is false  (for n ≥ 2), that is not the mistaken reasoning. The real flaw in the proof is that the induction step is valid for a “typical”  value of n, say, n = 3. The flaw, however, is in the induction step when n = 1. In this case, for n+1 = 2  horses, there are no “middle” horses, this makes the argument to collapse.

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.