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:

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31 Comments on “Performing SURDgery”

  1. Prathamesh Khairnar Says:

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

    • CyberChalky Says:

      Yes – if you want a bonus point.

      • Prathamesh Khairnar Says:

        OHH MAAA GODD..okay sir i read how to do the square root algorithm πŸ˜€

        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 😦

        anyways i learnt the new method i will show u in the next class if thats alright πŸ˜›

  2. Bernice Bagano Says:

    Hi Sir,

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

  3. MILICA Says:

    hey sir how do you work out
    -√5*2√8/5√10+7
    ???

    • Prathamesh Khairnar Says:

      first do the multiplication

      (-√5 X 2√8)
      = ———–
      (5√10+7)

      -2√40
      = ——–
      5√10 + 7

      -4√10
      = ——-
      5√10 + 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 πŸ™‚

      • Prathamesh Khairnar Says:

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

      • Prathamesh Khairnar Says:

        i will do it again sorry…

        (-√5 X 2√8)
        = ——-
        (5√10+7)

        -2√40
        = ——–—
        5√10 + 7

        -4√10
        = β€”β€”—-
        5√10 + 7

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

        so we get:

        -4√10 5√10-7
        =—— X ——
        5√10 + 7 5√10-7

        -20√100 + 28√10
        = —————-
        25√100 – 49

        -200+28√10
        = ———-
        250-49

        -200+28√10
        = ————-
        201

        we cant simplify anyfurther.. 😦

      • Ellayne Garcia Says:

        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 5√10 + 7. So I was thinking that

        -2√40 / 5√10+7 would be written down as:

        -2√40
        —– + 7
        5√10

        instead of

        -2√40
        —— which actually means (-2√40)/(5√10+7)
        5√10+7

        The first equation would then result to 6.2.

        Is that right, Sir?

      • Ellayne Garcia Says:

        Exercise 4.1
        1. k) -√5 x 2√8 / 5√10 + 7

  4. Nama Kadrie Says:

    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?

  5. Bernice Bagano Says:

    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.

    2√40
    = ———– [this is when we use rationalising]
    5 + √10

    2√40
    = ————- [and this is when we don’t]
    5 √10

    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) πŸ˜€

    • Ellayne Garcia Says:

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

      -2√40
      ——- [which is Bernice’s point]
      5√10 + 7

      He added 7 to the denominator [5√10] instead of adding it to [-2√40 / 5√10] which I think made the answer wrong.

  6. prathamesh Says:

    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

  7. prathamesh Says:

    dude bernice but its two terms in the denomitor
    sorry i am on phone i cant use the under root sign 😦

    The two terms are:
    5 (root) 10
    And
    +7

  8. Lauren Tampaline Says:

    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 πŸ™‚

    • Bernice Bagano Says:

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

      • CyberChalky Says:

        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…

    • Nama Kadrie Says:

      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?

    • CyberChalky Says:

      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.

  9. Prathamesh Khairnar Says:

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

  10. prathamesh Says:

    yeah sir, i am kinda obessed with the colour blue πŸ™‚


  11. […] 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 […]


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