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**Toast****Real Member**- Registered: 2006-10-08
- Posts: 1,321

To complete these questions it will be necessary to know the Index Laws:

Law 1:

Law 2:

Law 3:

Law 4:

Law 5:

And the rational index laws:

Also, we define

, whereTo prove this:

, but

1: Simplify with positive indices:

a)

*Last edited by Toast (2007-05-07 02:22:15)*

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**nova_angel****Member**- Registered: 2007-09-24
- Posts: 1

Hey..thats cool...may you post some level 9 or secondary 3(asia) practice to the forum... i need more practise...anyway thank you~~this is kinda cool...

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**kmlb123****Member**- Registered: 2007-12-28
- Posts: 1

ahhh...finally something i can understand ty for this stuff and could u upload some other stuff like medium level coordinate geometry (i drag at it) TY4 THIS

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 14,528

Hi kmlb123!

Happy to learn that you found this Exercise useful!

I shall post more exercises at Middle School/High School level;

certainly I shall try to post an exercise exclusively on Coordinate Geometry.

Character is who you are when no one is looking.

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**Ash Okas****Member**- Registered: 2013-01-07
- Posts: 1

Exercise online, good see the stuff, helpful for kids and all, It inspired me to join the lovely forum, evoked my forgotten love for maths

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

hi

Some people discover this forum but don't realise there is a huge wealth of excellent math teaching material including exercises at

Try it!

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,237

Hi Ash Okas;

Welcome to the forum. Tell us about yourself in "Introductions."

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi Toast! Nice set of problems!

Here is a variation of law 2 that is easy to use and always ends up with a non-negative exponent.

It takes care of all three cases: m>n, m=n, m<n. It is especially nice when the problem involves

negative exponents.

(p is the opposite of the smaller of m and n) For example recalling that a^0 = 1 we get

( Whether this is true for a=0 has been thoroughly explored in other threads of this forum.)

m n p a^(m+p)/a^(n+p)

2 5 -2 a^0 / a^3 = 1/(a^3)

3 -2 2 a^5 / a^0 = a^5

-5 -3 5 a^0 / a^2 = 1/(a^2)

4 -7 7 a^11 / a^0 = a^11

3 3 -3 a^0 / a^0 = 1

-4 -4 4 a^0 / a^0 = 1

Of course the last two cases here would obviously be 1 from the start.

This law takes care of all cases for positive, negative or zero exponents m and n, leaving the

answer with a non-negative exponent for a in the numerator or denominator as most books

require for the answer.

If p is anything else but -min(m,n) the equality is still true, but it will not be in "simplified" form.

Play around with it a bit and I think you will find it is quite handy and easy to use.

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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