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#4

(Original post by

If (a^2 + b^2)^1/2 = c then why doesn't a + b = c ?

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**maths_4_life**)If (a^2 + b^2)^1/2 = c then why doesn't a + b = c ?

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#5

From your first identity

From your second identity

Which would imply for both of your identities to be equal, a or b must equal 0. Otherwise they are never equal

From your second identity

Which would imply for both of your identities to be equal, a or b must equal 0. Otherwise they are never equal

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#6

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#7

oh ok

(Original post by

I'm assuming both statement are equivalent to show that a contradiction arrives from this equivalency, which is 2ab=0, which cannot be true unless a and or b =0,

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**drandy76**)I'm assuming both statement are equivalent to show that a contradiction arrives from this equivalency, which is 2ab=0, which cannot be true unless a and or b =0,

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#8

**maths_4_life**)

If (a^2 + b^2)^1/2 = c then why doesn't a + b = c ?

Posted from TSR Mobile

*linearity.*I'm going to write in a very informal and unrigorous way that is meant to supplement your intuition, so nobody hate on me.

Okay - first off, try and drill the practice of checking special cases in your head as you write something down. So if you're writing try and practice to get your head to test the case for example.

It's fairly obvious that right away this gives you but it is not the case that so that should set off alarm bells right away.

Okay, now onto me waffling about linearity: so things are linear when they satisfy (amongst other things) the fact that , it's quite a privilege to work with linear things, it makes life very easy.

A few examples of linear things is, well... a straight line, is linear because , makes life very easy, you don't need to worry about stuff so much.

Another example of linear things is integration or differentiation, although we're moving away from the realm of functions and into the realm of operators here (this needn't concern us, though) we can call differentiation linear because - see, life is easy! We don't need to worry about whether we add then differentiate or differentiate and add, meh, it all works out to be the same thing.

Okay, linear is easy... that kinda makes nearly everything not being linear kinda make sense, because when is maths easy, eh? Functions, in general aren't linear. isn't a linear function, unfortunately. I've just said that square rooting isn't a linear function! So it

*does not*hold that .

In general, square rooting is a messy business, it works from (we're constraining ourself to the reals here, it we want to go complex, bleurgh), we need to take care of the signs, restrict our domains, makes sure to use moduli bars, etc... so it hardly stands to reason that square rooting is going to be

*linear*and make life easy for us, oh no, that's not what little Mr. Square Root is going to do. He's going to stand there and make life hard.

Square rooting and logarithiming (made that word up, hi) kind of provoke the same reaction in me, whenever I see or I know right away I can split that up into or and make life easy, but as soon as I see or , I close my eyes and go cry a little bit because

*there's nothing I can really do now!*

So, hopefully, that mildly entertaining drivel has let you intuit that .

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#11

(Original post by

Dying.

**Zacken**)Dying.

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#12

(Original post by

x.

**Zacken**)x.

**and the absolute value of −3 is also 3**.<---- i don't understand this bit can you explain pls?

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#13

(Original post by

the absolute value of 3 is 3,

**thefatone**)the absolute value of 3 is 3,

**and the absolute value of −3 is also 3**.<---- i don't understand this bit can you explain pls?Posted from TSR Mobile

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#14

(Original post by

I like to think of the absolute value as the distance from the origin

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**drandy76**)I like to think of the absolute value as the distance from the origin

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#15

(Original post by

i don't quite understand that either....

**thefatone**)i don't quite understand that either....

3 is also 3 units away from 0

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#16

**thefatone**)

the absolute value of 3 is 3,

**and the absolute value of −3 is also 3**.<---- i don't understand this bit can you explain pls?

So plug in into the function we get that since so we take the top bit of the definition:

Plug in into the function we get that since so we take the bottom bit of the definition .

Essentially, the absolute value functions strips away the sign of the number and makes it positive.

Because the absolute value function is defined piecewise, it stands to reason why it is not differentiable at its cusp, more specifically: so the derivative is undefined for and that is because it is undefined there.

Most of this post is useless extra knowledge, but... meh.

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#18

(Original post by

We define the absolute value function:

So plug in into the function we get that since so we take the top bit of the definition:

Plug in into the function we get that since so we take the bottom bit of the definition .

Essentially, the absolute value functions strips away the sign of the number and makes it positive.

Because the absolute value function is defined piecewise, it stands to reason why it is not differentiable at its cusp, more specifically: so the derivative is undefined for and that is because it is undefined there.

Most of this post is useless extra knowledge, but... meh.

**Zacken**)We define the absolute value function:

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.

`\displaystyle`

\begin{equation*} |x| = \begin{cases} x \quad \text{for } \, x \geq 0 \\ -x \quad \text{for } \, x < 0\end{cases}\end{equation{*}

\begin{equation*} |x| = \begin{cases} x \quad \text{for } \, x \geq 0 \\ -x \quad \text{for } \, x < 0\end{cases}\end{equation{*}

So plug in into the function we get that since so we take the top bit of the definition:

Plug in into the function we get that since so we take the bottom bit of the definition .

Essentially, the absolute value functions strips away the sign of the number and makes it positive.

Because the absolute value function is defined piecewise, it stands to reason why it is not differentiable at its cusp, more specifically: so the derivative is undefined for and that is because it is undefined there.

Most of this post is useless extra knowledge, but... meh.

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#20

(Original post by

I wouldn't put it that way - Zain explains it well

**Student403**)I wouldn't put it that way - Zain explains it well

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