A pythagorean conundrum

[ZeroZero]

I am currently studying Turing machines and I have set myself the following problem

 

 

 

Using Pythagorian theory

 

1

Adjacent side = 12 inches

 

Opposite = 12 inches

 

Therefore hypononuse =19.7 (approx)

 

 

2

 

Adjacent = 1 foot

 

Opposite = 1 foot

 

 

 

therefore hypotonuse = square root of 2 which is a non computable 'irrational number'

 

Same triangle - same dimensions

 

First example - computable, second non computable - but the are the same.

 

 

Go easy on me - my maths is not great... simple explanation appreciated.

 

 

 

Zero

Edited May 3, 2012 by ZeroZero

[Joatmon]

The 12 inch triangle has a hypotenuse of approx 16.97 which is twelve times the square root of two which is irrational.

Edited May 3, 2012 by Joatmon

[michel123456]

That's the beauty

 

If you take a circle

1. say the radius is 1, then the perimeter (2Pi) is irrational

or

2. say the perimeter is 1, then the radius (1/2Pi) is irrational.

 

There is no absolute irrational, there is only irrationality relative to something. Pi is irrational relatively to number one AKA it cannot be described by a fraction. And as you must know, you need 2 numbers at least to make a fraction.

 

For example, you can cut a string and decide arbitrarily the the length of the string will be Pi in some new weird unit system. Then curve the string in a circle, find the diameter: it is the number one in your weird unit system.

It's all relative.

 

Does that help?

Edited May 3, 2012 by michel123456

[ZeroZero]

Sorry Michel not yet - it has not clicked in my irrational mind

 

 

The bit about there being no absolute irrational numbers only relative seems a bit of a clue

[Visionem Ex Illuminatio]

Hey ZeroZero

 

Let's take a look at what you said... (if you find the math below intimidating, don't worry; it always looks scary when it actually represents a simple geometric situation)

 

• The first right triangle has legs of 12 units and 12 units.

• The second right triangle has legs of 1 unit and 1 unit.

Now, these triangles are not the same, as you said. They do not have the exact same dimensions. But, they are similar, meaning their dimensions are proportional. You're implying that they're congruent, which in this case would result in [math]12=1[/math] D:

 

There's a theorem in geometry that says that if two triangles have two similar legs and one congruent angle, then the triangles are similar. In other words, all you have to know are the 2 legs and the 90 degree angle and if those are proportional, then the triangles are also proportional (you don't need to know the other two angles or the length of the hypotenuse; that's it!)

 

[math]a^2+b^2=c^2[/math] is the Pythagorean Theorem, as you probably know. However, since you've picked the interesting property that the legs are equal to each other, we can make this:

 

[math]a^2+a^2=c^2[/math] since [math]a[/math] equals [math]b[/math] in this case. Then we further simplify to [math]2a^2=c^2[/math].

Now here's the important part...

 

Look at [math]2a^2=c^2[/math]. We can take the square root of both sides and make it [math]\sqrt{2a^2}=\sqrt{c^2}[/math].

 

On the right side, it would just simplify to [math]c[/math], since the square root of a number is squared is just that number.

 

On the left, with a little algebra rule, we know that [math]\sqrt{2a^2}[/math] is the same as [math]\sqrt{2}\times\sqrt{a^2}[/math]. (again, the [math]\sqrt{a^2}[/math] just becomes [math]a[/math])

 

So now! We're left with [math]a\sqrt{2}=c[/math]. Remember, [math]c[/math] is the length of the hypotenuse. So basically, [math]c[/math] is just equal to the length of one of the legs times the square root of 2. [math]\sqrt{2}[/math] is irrational, and any multiple of it is also irrational. So no matter what integers you have for the length of the legs, the answer will be irrational.

 

The first triangle you showed has an irrational hypotenuse, just like the second one. It's not that the first was fine and computable while the second one wasn't, both of them are irrational and "not computable" in that way. The 12-12-? triangle has a hypotenuse of [math]12\times\sqrt{2}[/math] (as seen from the simpler equation we made). Yes, it's approximately 16.97056...etc. However, [math]\sqrt{2}[/math] is no problem either; it's approximately 1.414.

 

Don't worry, there's no Pythagorean conundrum.

 

Sorry if this was much too long, but I felt that too many people get confused with algebra on the square root part, so... hoped this help!

Edited May 3, 2012 by Visionem Ex Illuminatio

[doG]

They do not have the exact same dimensions....

Actually they do since 12 inches equals 1 foot. They are identical triangles where only the units of measure have changed. The OP is claiming that the hypotenuse of the first is rational and the second irrational.

[ZeroZero]

So 12 X 12 is an irrational number?

 

 

 

And yes 12 inches =1 foot the triangles are identical in all respects not merely congruent. The only variation is the arbitrary definition of measurement units. This would imply that we only get irrational numbers if we use certain arbitrary divisions working in metres would produce different results from imperial.

[Greg H.]

Actually both numbers are irrational.

 

The problem is you're trying to compare two numbers with different units of measure and in both science and math, that will confuse you every time.

 

Ex 1

12² + 12² = c²

144 + 144 = c²

288 = c ²

[math]\sqrt{288}[/math] = c

[math]\sqrt{144}\times\sqrt{2}[/math] = c

[math]12\sqrt{2}[/math] = c <- This is in inches. Remember this, it will be important later.

 

Ex2

1² + 1² = c²

1 + 1 = c²

2 = c²

[math]\sqrt{2}[/math] = c <- This is in feet.

 

To convert them to be the same unit, you either multiply ex 2 by 12 inches/foot or you divide Ex 1 by 12 inches per 1 foot

 

In either case, the answers are both the same -> [math]\sqrt{2}[/math] feet (aka [math]12\sqrt{2}[/math] inches) and they are both irrational.

 

HTH ZZ

 

Edit: I learned to do the math thing so it's easier to read.

Edited May 3, 2012 by Greg H.

[Visionem Ex Illuminatio]

Wow, terrible misconception. I thought you meant 12 inches vs 1 inch. My fault; should've read closer. Well, Greg H. made a good explanation.

 

So 12 x 12 is an irrational number?

 

Where did you get 12 x 12 from? I said the relationship between the hypotenuse length ( c ) and the leg length (a) is [math]a\times\sqrt{2}=c[/math]. In this case' date=' it's 12 times sqrt(2), which as others have explained is just as irrational as sqrt(2).

 

Both are, in the way you stated, "non-computable", and require approximations. [math']12\times\sqrt{2}[/math] in inches is the same as [math]\sqrt{2}[/math] in feet... as you see, the contrasting coefficients are 12 and 1. Makes sense

Edited May 3, 2012 by Visionem Ex Illuminatio

[Joatmon]

Another similar triangle could be considered with the two sides each being the square root of two. The hypotenuse would then be the rational number two.

(IMO) Any triangle that is similar to the one described by the OP will have at least one side that is irrational. Sometimes two sides and sometimes all three sides will be irrational.

Edited May 4, 2012 by Joatmon

[ewmon]

Adjacent side = 12 inches

 

Opposite = 12 inches

 

Therefore hypononuse =19.7 (approx)

Hey ZeroZero,

 

If any of this is still not registering, you should tell us how you arrived at "19.7 (approx)" in the above problem. It looks like you added 12² plus 12² (144 + 144) and got 388 by mistake, and then approximated its square root. Don't worry, taking the square root of this wrong number still serves our purpose here. The thing is, you could only approximate the square root ... because it's irrational (√388 = 2√97, and √97 is irrational). The same thing happens with 12² + 12² → √288 = 12√2, and √2 is irrational.

 

PS — If you're having difficulty spelling "hypotenuse", just think of it as hypo + tense, or under tension (imagine a cord stretched tight along the diagonal), and the "u" naturally finds its way into the word from our pronunciation of it.

[Joatmon]

I think another way at looking at irrationality, which I think in essence is similar to Michel's idea of relative ratios is this:-

 

Take a ruler divided in to 12 equal sections and number the sections 1 to 12. e.g. a normal 12 inch ruler. Now ask someone to mark where the square root of two is and it is impossible. Whatever mark they make will represent either more or less than the square root of two.

 

Take the same ruler but this time number the sections 1*root 2 to 12*root 2. Now ask someone to mark where 1 is and that is impossible. Whatever mark they make will represent either more or less than 1.

 

If I've misunderstood Michel's point then apologies in advance!

Edited May 4, 2012 by Joatmon

[ZeroZero]

Yes I made two schoholboy mistakes.

 

1] 12*12 + 12+12 = 288 not 388

 

2] I used a caculator to get the square root of 288 not a good move I think

 

 

 

 

so the square root of 288 is an irrational number too?

 

 

If i were to use other arbitrary measures to divide the lengths would I also get irrational numbers?

 

Are all answers to the pythagorean calculation of the length of the hypotonuse irrational (expect perhaps when the opposite and adjacent are irrational - now there is an interesting question)

 

I dont know how to indentify an irrational number except to say it is not a product of an integer fraction - including re-occurring such as 1/3

[doG]

so the square root of 288 is an irrational number too?

Yes, since it equals [math]12\sqrt{2}[/math]

 

If i were to use other arbitrary measures to divide the lengths would I also get irrational numbers?

For that particular triangle you could choose units that make the hypotenuse a nice even square but it could make your sides irrational.

 

Are all answers to the pythagorean calculation of the length of the hypotonuse irrational (expect perhaps when the opposite and adjacent are irrational - now there is an interesting question)

A triangle with sides of 3 and 4 units in length has a hypotenuse of 5 units. All 3 sides are integers and therefore rational.

[Bignose]

Are all answers to the pythagorean calculation of the length of the hypotonuse irrational (expect perhaps when the opposite and adjacent are irrational - now there is an interesting question)

 

In addition to the 3-4-5 triangle listed in the post above me, the other favorite one of math classes is the 5-12-13 triangle. Sets of three numbers that can make integer sided right triangles are known as Pythagorean Triples. Here's a list with a bunch of them. http://www.tsm-resources.com/alists/trip.html As you can imagine, there is an infinite number of them, especially considering multiples of the known ones. That is, since 3-4-5 is a Pythagorean Triple, so is 6-8-10.

 

That link above also shows how one can always generate a Pythagorean triple. Take 2 integers, m & n, and let m>n. Then 2mn, m²+n², m²-n² will always be a Pythagorean Triple. So long as you start with m & n as integers, all three of those calculations will also result in integers.

Edited May 6, 2012 by Bignose

[Greg H.]

I dont know how to indentify an irrational number except to say it is not a product of an integer fraction - including re-occurring such as 1/3

 

That's exactly what an irrational number is.

[ZeroZero]

Thanks for your help all. I think I get this now. All right angled triangles of the form 45,45, 90 degrees have irrational numbers as the hypotonuse. Others don't - necessesarily (though some may have).

 

 

 

many thanks