How many ways are there to prove the Pythagorean theorem Betty Fei

What do Euclid,

twelve-year-old Einstein,

and American President James Garfield
have in common?

They all came up with elegant
proofs for the famous Pythagorean theorem,

the rule that says for a right triangle,

the square of one side plus
the square of the other side

is equal to the square of the hypotenuse.

In other words, a²+b²=c².

This statement is one of the most
fundamental rules of geometry,

and the basis for practical applications,

like constructing stable buildings
and triangulating GPS coordinates.

The theorem is named for Pythagoras,

a Greek philosopher and mathematician
in the 6th century B.C.,

but it was known more than a
thousand years earlier.

A Babylonian tablet from around 1800 B.C.
lists 15 sets of numbers

that satisfy the theorem.

Some historians speculate
that Ancient Egyptian surveyors

used one such set of numbers, 3, 4, 5,
to make square corners.

The theory is that surveyors could stretch
a knotted rope with twelve equal segments

to form a triangle with sides of length
3, 4 and 5.

According to the converse
of the Pythagorean theorem,

that has to make a right triangle,

and, therefore, a square corner.

And the earliest known
Indian mathematical texts

written between 800 and 600 B.C.

state that a rope stretched across
the diagonal of a square

produces a square twice as large
as the original one.

That relationship can be derived
from the Pythagorean theorem.

But how do we know
that the theorem is true

for every right triangle
on a flat surface,

not just the ones these mathematicians
and surveyors knew about?

Because we can prove it.

Proofs use existing mathematical rules
and logic

to demonstrate that a theorem
must hold true all the time.

One classic proof often attributed
to Pythagoras himself

uses a strategy called
proof by rearrangement.

Take four identical right triangles
with side lengths a and b

and hypotenuse length c.

Arrange them so that their hypotenuses
form a tilted square.

The area of that square is c².

Now rearrange the triangles
into two rectangles,

leaving smaller squares on either side.

The areas of those squares
are a² and b².

Here’s the key.

The total area of
the figure didn’t change,

and the areas of the triangles
didn’t change.

So the empty space in one, c²

must be equal to
the empty space in the other,

a² + b².

Another proof comes from a fellow Greek
mathematician Euclid

and was also stumbled upon
almost 2,000 years later

by twelve-year-old Einstein.

This proof divides one right triangle
into two others

and uses the principle that if the
corresponding angles of two triangles

are the same,

the ratio of their sides
is the same, too.

So for these three similar triangles,

you can write these expressions
for their sides.

Next, rearrange the terms.

And finally, add the two equations
together and simplify to get

ab²+ac²=bc²,

or a²+b²=c².

Here’s one that uses tessellation,

a repeating geometric pattern
for a more visual proof.

Can you see how it works?

Pause the video if you’d like some time
to think about it.

Here’s the answer.

The dark gray square is a²

and the light gray one is b².

The one outlined in blue is c².

Each blue outlined square
contains the pieces of exactly one dark

and one light gray square,

proving the Pythagorean theorem again.

And if you’d really like
to convince yourself,

you could build a turntable
with three square boxes of equal depth

connected to each other
around a right triangle.

If you fill the largest square with water
and spin the turntable,

the water from the large square
will perfectly fill the two smaller ones.

The Pythagorean theorem has more
than 350 proofs, and counting,

ranging from brilliant to obscure.

Can you add your own to the mix?

欧几里得、

12 岁的爱因斯坦

和美国总统詹姆斯·加菲尔德
有什么共同点?

他们都
为著名的勾股定理提出了优雅的证明,

即对于直角三角形,

一边
的平方加上另一边

的平方等于斜边的平方。

换句话说,a²+b²=c²。

该语句是
几何最基本的规则之一,

也是实际应用的基础,

例如构建稳定的建筑物
和三角测量 GPS 坐标。

该定理以公元前 6 世纪

的希腊哲学家和数学家
毕达哥拉斯命名,

但早在一千多年前就已为人所知

公元前 1800 年左右的巴比伦石板
列出

满足定理的 15 组数字。

一些历史学家推测
,古埃及测量员

使用一组这样的数字 3、4、5
来制作方角。

该理论是,测量员可以将
一根打结的绳索拉伸成 12 个相等的段

,形成一个边长为
3、4 和 5 的三角形。

根据
勾股定理的逆,

它必须形成一个直角三角形

,因此, 方角。

以及已知最早的
印度数学

著作,写于公元前 800 年至 600 年之间。

陈述一条
穿过正方形对角线的绳索

产生的正方形
是原来的两倍大。

这种关系可以
从勾股定理推导出来。

但是我们怎么知道
这个定理

对于平面上的每个直角三角形都是正确的

而不仅仅是这些数学家
和测量员知道的那些呢?

因为我们可以证明。

证明使用现有的数学规则
和逻辑

来证明定理
必须始终成立。

一个经常归因
于毕达哥拉斯本人的经典证明

使用了一种称为
重排证明的策略。

取四个相同的直角三角形
,边长为 a 和 b

,斜边长为 c。

排列它们,使它们的斜边
形成一个倾斜的正方形。

那个正方形的面积是c²。

现在将三角形重新
排列成两个矩形,

在两边留下较小的正方形。

这些正方形的面积
是a²和b²。

这是关键。 图形

的总
面积没有变化

,三角形的面积
也没有变化。

因此,一个空间 c² 中的空白空间

必须等于
另一个空间

a² + b² 中的空白空间。

另一个证明来自一位希腊
数学家欧几里得

,也是在
将近 2000 年

后被 12 岁的爱因斯坦偶然发现的。

这个证明将一个直角三角形
分成两个其他三角形,

并使用如果
两个三角形的对应角

相同,

则它们的边之比
也相同的原理。

所以对于这三个相似的三角形,

你可以
为它们的边写这些表达式。

接下来,重新排列术语。

最后,将这两个方程加
在一起并化简为

a²+ac²=bc²,

或 a²+b²=c²。

这是使用镶嵌的一个,

一种重复的几何图案,
用于更直观的证明。

你能看到它是如何工作的吗?

如果您想
花点时间考虑一下,请暂停视频。

这是答案。

深灰色方块是a²

,浅灰色方块是b²。

蓝色框出的是c²。

每个蓝色轮廓的正方形都
包含一个黑色

正方形和一个浅灰色正方形的碎片,

再次证明了勾股定理。

如果你真的
想说服自己,

你可以构建一个转盘
,三个正方形的盒子

围绕一个直角三角形相互连接。

如果你用水填充最大的正方形
并旋转转盘,

大正方形的水
将完美地填充两个较小的正方形。

毕达哥拉斯定理有
350 多个证明,而且数不胜数,

从辉煌到晦涩。

您可以添加自己的组合吗?