## Performing SURDgery

Everyone knows about numbers. How difficult can they be, really?

You start with one, and go up – it’s easy.

Of course, there is zero, and you can go down into the negative numbers. That’s still OK

What about between numbers – fractions and decimals. No problems yet, right?

OK, so what else can happen –Β well we start divding the numbers into types. Natural (counting), Whole (natural and zero), Integers (Whole and negative), Rational (numbers that can be expressed as fractions). You can calculate the fraction of any repeating decimal by a simple process (shown at either of these links: 1,2)

That about does it, right?

Well, no – there are the numbers that can’t be written as a fraction (the ratio of two numbers). These are non repeating, non terminating decimals. Things like the square root of two, which has been calculated to one **trillion **digits with no patterns found.

But it doesn’t matter how far you go, there will **never** be a pattern. We know this, because we can prove it. Here is a page of proofs – they are quite scary, so you might not want to look.

There is a way to actually find the decimal approximation (to any degree of accuracy) by using an algorithm. If you can use this algorithm, you are demonstrating a high level of ability with arithmetic, algebra and mathematical processes. As such anyone who can show me a handworked example of the calculation of the square root of a prime number (must not be the same as anyone in your class!!) of your choice will earn a bonus point.

While all of the above is interesting, you need to learn is the simplification of surds. Fortunately, I’ve already got a lot of information for you at a previous post!

Also you should check out these videos – they will supplement our class instruction and your textbook. Here is the first one:

**Explore posts in the same categories:**Mathematics

February 21, 2011 at 2:48 PM

okay sir, are we meant to show u an example (working out) of an estimation of a surd?

February 22, 2011 at 7:01 PM

Yes – if you want a bonus point.

February 22, 2011 at 8:23 PM

OHH MAAA GODD..okay sir i read how to do the square root algorithm π

this is exactly how i do it but someone managed to put it in a equation lol

now i checked the example i showed u in class for some reason( i have no idea why) but i did a miss-calculation π¦

anyways i learnt the new method i will show u in the next class if thats alright π

February 23, 2011 at 4:08 PM

Hi Sir,

It’s about the bound reference, will it matter if we use a pencil or pen?

February 26, 2011 at 3:56 PM

Pen is better as it doesn’t fade/ rub away.

February 23, 2011 at 7:55 PM

hey sir how do you work out

-β5*2β8/5β10+7

???

February 26, 2011 at 5:46 PM

first do the multiplication

(-β5 X 2β8)

= ———–

(5β10+7)

-2β40

= ——–

5β10 + 7

-4β10

= ——-

5β10 + 7

= – 0.8 + 7

= 6.2

i might be wrong though, unless if i am right.

and yes that sentence totally made sense 0.o π

March 2, 2011 at 5:48 PM

OH MY GOD I AM WORNG!!!! *shoots myself*

March 2, 2011 at 6:00 PM

i will do it again sorry…

(-β5 X 2β8)

= ——-

(5β10+7)

-2β40

= βββ—

5β10 + 7

-4β10

= ββ—-

5β10 + 7

AFTER THIS STEP WE HAVE TO RATIONALISE THE DENOMINATOR (this is where i went wrong)

so we get:

-4β10 5β10-7

=—— X ——

5β10 + 7 5β10-7

-20β100 + 28β10

= —————-

25β100 – 49

-200+28β10

= ———-

250-49

-200+28β10

= ————-

201

we cant simplify anyfurther.. π¦

March 2, 2011 at 7:11 PM

Hi Sir,

I got 6.2 on this question and I think that what Prathamesh did on the first question was the right one. I thought that there was no need to rationalise the denominator since there was no parenthesis enclosing 5β10 + 7. So I was thinking that

-2β40 / 5β10+7 would be written down as:

-2β40

—– + 7

5β10

instead of

-2β40

—— which actually means (-2β40)/(5β10+7)

5β10+7

The first equation would then result to 6.2.

Is that right, Sir?

March 2, 2011 at 8:22 PM

Which question is it from the textbook? The correct answer depends on whether the denominator is 5β10 or 5β10+7.

March 2, 2011 at 9:00 PM

Exercise 4.1

1. k) -β5 x 2β8 / 5β10 + 7

March 1, 2011 at 11:38 PM

Heron’s Formula =

S (third variable) = a+b+c

_______

2

Once you have found out “S”

Area of you triangle is going to be equal to :

Area:Square root of “S” Times it by (s-a) (s-b) (s-c)

Is that right, Sir?

March 5, 2011 at 9:26 PM

That’s the one, Nama.

You might be interested in looking up Brahmagupta’s formula – it takes it a step further.

March 2, 2011 at 9:33 PM

Prathu,

You already got it the first time you tried solving the question/equation.

I’m not entirely sure about this but I think that we only use rationalising whenever + and – actually is written down in the question.

e.g.

2β40

= ———– [this is when we use rationalising]

5 + β10

2β40

= ————- [and this is when we don’t]

5 β10

I’m not sure if this will make sense to you, but I think that’s how it works (unless there’s a rule as to why it works this way) π

March 2, 2011 at 9:44 PM

I totally agree with that. The only reason he rationalised the denominator is because he did it this way:

-2β40

——- [which is Bernice’s point]

5β10 + 7

He added 7 to the denominator [5β10] instead of adding it to [-2β40 / 5β10] which I think made the answer wrong.

March 3, 2011 at 6:48 AM

its doesnt matter we have to rationalise the denominator even if it is + or –

try is out ur self… Wat ever it is… + and – both are going to give u the middle term

(a+b)(a+b)= a^2+2ab+b^2

And..

(a-b)(a-b)= a^2-2ab-b^2

So we have to rationalise the denominator

March 5, 2011 at 8:47 AM

Always use the opposite sign to the denominator!

March 3, 2011 at 6:50 AM

dude bernice but its two terms in the denomitor

sorry i am on phone i cant use the under root sign π¦

The two terms are:

5 (root) 10

And

+7

March 4, 2011 at 4:29 PM

Hey Sir,

I worked out the origami thing we made the other day (sorry, i forgot what it’s called). It has a total external surface area of 3234.4cm squared, has 80 external faces, 100 vertices (internal and external) and I’m pretty sure it has 130 edges (internal and external). By the way, i thought it was a really cool activity- better than textbook work. So on behalf of 10H, thankyou π

March 4, 2011 at 6:31 PM

Lauren,

I think it’s called a icosagon? (Haha, I searched it up, because I forgot as well). AGREED! Thanks Sir! π

March 5, 2011 at 8:52 AM

Good try Bernice. You are *almost* right

An Icosagon is a twenty *sided* two dimensional shape.

An Icosahedron is a twenty *faced* 3-d object.

The shape we made has more than twenty faces, as each face has a pyramid on it…

March 5, 2011 at 3:39 AM

Hey

Lauren,

I am pretty sure, your vertices are not correct. They do not add up to 100.

Try again, you might get it right.

BTW, did you use Euler’s rule to solve this equation?

March 5, 2011 at 8:50 AM

Hi Lauren,

I’m glad you enjoyed it – it is fun for me too!

Your answers are a little off, remember that it is a “Greater stellated icosahedron”. It might also help to look up Euler’s Rule.

March 5, 2011 at 1:40 PM

THE ICOSAHEDRON!!! http://img806.imageshack.us/i/dsc02189k.jpg/

March 5, 2011 at 5:06 PM

Congrats Prathesh; The winner of the Greater Stellated Icosahedron

Wow. That’s a *blue* wall!

March 5, 2011 at 5:13 PM

Lucky you π Congratulations!

March 5, 2011 at 5:19 PM

Wow! π Congrats!

Sir, do we still need to work the questions out and hand it to you?

March 5, 2011 at 9:25 PM

Yes, you have to show that you understand the principle, and specifically Euler’s Rule.

March 5, 2011 at 6:28 PM

yeah sir, i am kinda obessed with the colour blue π

March 14, 2011 at 10:47 AM

[…] from flat patterns – we made the greater stellated icosahedron – which was won by Prathu. But for the purposes of our class, we need to be able to calculate the total surface area of a […]