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

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?

(edited 7 years ago)

Reply 1

B_9710

\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.

(edited 7 years ago)

Reply 2

NotNotBatman

index laws.

a^m \times a^n = a^{m+n}

, you have to add the powers not multiply.

Reply 3

Notnek

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.

(edited 7 years ago)

Reply 4

RDKGames

Study Forum Helper

Original post by Athematica

\sqrt{a} = a^{1/2}

\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.

(edited 7 years ago)

Reply 5

Athematica

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.

(edited 7 years ago)

Reply 6

RDKGames

Study Forum Helper

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. :wink:

Spoiler

Reply 7

Athematica

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

Reply 8

NotNotBatman

Original post by Athematica

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

I was just about to respond to that. Days like this happen sometimes.

Reply 9

Athematica

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

That is the correct answer :tongue:

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

Reply 10

Zacken

Original post by Athematica

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

That is the correct answer :tongue:

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

Reply 11

Athematica

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

Reply 12

RDKGames

Study Forum Helper

Original post by Athematica

[video="youtube;u7Z9UnWOJNY"]https://www.youtube.com/watch?v=u7Z9UnWOJNY[/video]

Reply 13

Athematica

Haha. Great channel! :biggrin:

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?