Join TSR now and get all your revision questions answeredSign up now

Why does a / root(a) = root(a) ?

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    we have

    3 / root(3)

    A.

    which is identical to

    (3^1) / (3^1/2)

    which is identical to

    (3^1) x (3^-1/2)

    as

    1 x -(1/2) = -(1/2)

    it follows that

    (3^1) x (3^-1/2) = 3^(-1/2)

    which is the same as

    1 / (3^1/2)

    B.

    3 / root(3) x root(3) / root(3) = 3 *root(3) / root(3)^2

    which is equal to

    3*root(3) / 3

    .

    but

    root(3) does not equal 1/root(3).

    Have I misunderstood something?
    Online

    3
    ReputationRep:
     \frac{3\sqrt 3}{3} = \sqrt 3 .
    The 3 on top and bottom cancel.
    I'm not sure if I have read the question right.
    Offline

    3
    ReputationRep:
    index laws. a^m \times a^n = a^{m+n}, you have to add the powers not multiply.
    Offline

    3
    ReputationRep:
    (Original post by Athematica)
    (3^1) x (3^-1/2)

    it follows that

    (3^1) x (3^-1/2) = 3^(-1/2)
    I have marked in bold your first mistake.

    Think about the index laws : when you multiply two numbers with the same base, you add the powers, not multiply.
    Online

    3
    ReputationRep:
    (Original post by Athematica)
    we have

    3 / root(3)

    A.

    which is identical to

    (3^1) / (3^1/2)

    which is identical to

    (3^1) x (3^-1/2)

    as

    1 x -(1/2) = -(1/2)

    it follows that

    (3^1) x (3^-1/2) = 3^(-1/2)

    which is the same as

    1 / (3^1/2)

    B.

    3 / root(3) x root(3) / root(3) = 3 *root(3) / root(3)^2

    which is equal to

    3*root(3) / 3

    .

    but

    root(3) does not equal 1/root(3).

    Have I misunderstood something?
    \sqrt{a} = a^{1/2}

    So \displaystyle \frac{a}{\sqrt{a}}=\frac{a^1}{a^  {1/2}}=a^1 \cdot a^{-1/2}=a^{(1-1/2)}=a^{1/2}=\sqrt{a}

    (Original post by Athematica)
    Have I misunderstood something?
    Yeah only one of the most important basic rules concerning indices.
    • Thread Starter
    Offline

    2
    ReputationRep:
    It was a silly mistake to make. Thank you. It is coherent.

    (Original post by RDKGames)
    Yeah only one of the most important basic rules concerning indices.
    There is no need for snark. I made careless error.
    Online

    3
    ReputationRep:
    (Original post by Athematica)
    There is no need for snark. I made careless error.
    Live and learn, at least you won't make the same mistake again.
    Spoiler:
    Show
    I hope.
    • Thread Starter
    Offline

    2
    ReputationRep:
    I also made another error in typing when showing I understand that it works. This is not my evening.
    Offline

    3
    ReputationRep:
    (Original post by Athematica)
    I also made another error in typing when showing I understand that it works. This is not my evening.
    I was just about to respond to that. Days like this happen sometimes.
    • Thread Starter
    Offline

    2
    ReputationRep:
    a / root(a) = a^1 x a^(-1/2) = a^(1/2) = root(a)

    That is the correct answer

    I also learned through play with these that the series

    s = 1/root(1)-root(2) + 1/root(2)-root(3) + 1/root(3)-root(4) +1/root(4)-root(5) + ... + 1/root(99)-root(100)

    is s = 9, which is pretty cool
    Offline

    3
    ReputationRep:
    (Original post by Athematica)
    a / root(a) = a^1 x a^(-1/2) = a^(1/2) = root(a)

    That is the correct answer

    I also learned through play with these that the series

    s = 1/root(1)-root(2) + 1/root(2)-root(3) + 1/root(3)-root(4) +1/root(4)-root(5) + ... + 1/root(99)-root(100)

    is s = 9, which is pretty cool
    It's an example of a telescoping series once you rationalise
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    It's an example of a telescoping series once you rationalise
    Oh, I didn't realise this is a formal thing that happens.

    I noticed something similar when solving the paradox where you have a space to travel and can move at half the distance the step you did before, the first being half the distance of the space. You turn an infinite series into finite numbers which is quite satisfying
    Online

    3
    ReputationRep:
    (Original post by Athematica)
    Oh, I didn't realise this is a formal thing that happens.

    I noticed something similar when solving the paradox where you have a space to travel and can move at half the distance the step you did before, the first being half the distance of the space. You turn an infinite series into finite numbers which is quite satisfying
    • Thread Starter
    Offline

    2
    ReputationRep:
    Haha. Great channel!

    Yeah, that's the one :yep: Skimming through, they solve it the same way I did in class. Hadn't realised that is the usual way to do it so hi-five me?



    On another note. I am really, really bad at the maths challenge-y stuff my college are asking me to do before (October or November?) this year's senior one. Any tips?
    Online

    3
    ReputationRep:
    (Original post by Athematica)
    Haha. Great channel!

    Yeah, that's the one :yep: Skimming through, they solve it the same way I did in class. Hadn't realised that is the usual way to do it so hi-five me?
    If you're talking about doing something like S=... and then 2S=... then yeah it's a pretty standard method to do these sums. It's where \sum\limits_{r=1}^{n}r = \frac{1}{2}n(n+1) comes from. Gauss was the first person to use it, I think, at least for the sum of integers.
    Offline

    3
    ReputationRep:
    x^2/x=x
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Maker)
    x^2/x=x
    Omg this means we can use difference of two squares on surds!!!
 
 
 
Poll
Do you have exam superstitions?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.