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


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drandy76
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drandy76
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Should say provided a or b aren't 0 at the bottom


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thefatone
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(Original post by maths_4_life)
If (a^2 + b^2)^1/2 = c then why doesn't a + b = c ?


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^^ i think this is what you're assuming  \left( a^2 + b^2\right)^{\frac {1}{2}} \not = a+b
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Student403
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From your first identity

\sqrt {a^2+b^2} = c \\ a^2+b^2 = c^2

From your second identity

a+b=c \\ a^2+b^2 + 2ab = c^2

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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drandy76
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(Original post by thefatone)
^^ i think this is what you're assuming  \left( a^2 + b^2\right)^{\frac {1}{2}} \not = a+b
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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thefatone
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oh ok

(Original post by 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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Zacken
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(Original post by maths_4_life)
If (a^2 + b^2)^1/2 = c then why doesn't a + b = c ?


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The other users have provided you with perfectly valid proofs so I'm not going to go on with that, I'll give a few counter examples (remember, testing out special cases is always a good way of making sure what you're writing down is correct) and then go on and talk/waffle on a bit about something called 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 \sqrt{a^2 + b^2 } = c \Rightarrow a + b = c try and practice to get your head to test the case a=b=1 for example.

It's fairly obvious that right away this gives you \sqrt{a^2 + b^2} = \sqrt{2} but it is not the case that a+b = 2 =^? \sqrt{2} 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 f(x+y) = f(x) + f(y), 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, f(x) = ax is linear because f(x+y) = a(x+y) = ax + ay = f(x) + f(y), 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 \frac{\mathrm{d}}{\mathrm{d}x}(f(x) + g(x)) = \frac{\mathrm{d}}{\mathrm{d}x} f(x) + \frac{\mathrm{d}}{\mathrm{d}x} g(x) - 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. f(x) = \sqrt{x} isn't a linear function, unfortunately. I've just said that square rooting isn't a linear function! So it does not hold that \sqrt{a +b} = \sqrt{a} + \sqrt{b}.

In general, square rooting is a messy business, it works from \mathbb{R}^{+} \to \mathbb{R}^{+} (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 \sqrt{ab} or \ln xy I know right away I can split that up into \sqrt{a}\sqrt{b} or \ln x + \ln y and make life easy, but as soon as I see \sqrt{a+b} or \ln (x+y), 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 \sqrt{a^2 + b^2} \neq \sqrt{a^2} + \sqrt{b^2} = |a| + |b|.
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Student403
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(Original post by Zacken)
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Zacken
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(Original post by Student403)
...
Dying. :toofunny: :rofl:
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Student403
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(Original post by Zacken)
Dying. :toofunny: :rofl:
Dude I legitimately think you're better at maths than my teachers
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thefatone
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(Original post by Zacken)
x.
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?
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drandy76
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(Original post by 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?
I like to think of the absolute value as the distance from the origin


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thefatone
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(Original post by drandy76)
I like to think of the absolute value as the distance from the origin


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i don't quite understand that either....
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Student403
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(Original post by thefatone)
i don't quite understand that either....
-3 is 3 units away from 0
3 is also 3 units away from 0
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Zacken
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(Original post by 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?
We define the absolute value function:

\displaystyle

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

So plug in x=3 into the function |x| we get that since 3 &gt; 0 so we take the top bit of the definition: |3| = 3

Plug in x=-3 into the function |x| we get that since -3 &lt; 0 so we take the bottom bit of the definition |-3| = -(-3) = 3.

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: y = |x| \Rightarrow y^2 = x^2 \Rightarrow 2yy' = 2x \Rightarrow y' = \frac{x}{|x|} so the derivative is undefined for |x| =0 \iff x=0 and that is because it is undefined there.

Most of this post is useless extra knowledge, but... meh.
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thefatone
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(Original post by Student403)
-3 is 3 units away from 0
3 is also 3 units away from 0
oh that makes sense....

so ± are interchangeable?
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thefatone
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(Original post by 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 &lt; 0\end{cases}\end{equation{*}


So plug in x=3 into the function |x| we get that since 3 &gt; 0 so we take the top bit of the definition: |3| = 3

Plug in x=-3 into the function |x| we get that since -3 &lt; 0 so we take the bottom bit of the definition |-3| = -(-3) = 3.

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: y = |x| \Rightarrow y^2 = x^2 \Rightarrow 2yy' = 2x \Rightarrow y' = \frac{x}{|x|} so the derivative is undefined for |x| =0 \iff x=0 and that is because it is undefined there.

Most of this post is useless extra knowledge, but... meh.
seems pretty dangerous... do i learn this in c3 and c4?
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Student403
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(Original post by thefatone)
oh that makes sense....

so ± are interchangeable?
I wouldn't put it that way - Zain explains it well
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thefatone
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(Original post by Student403)
I wouldn't put it that way - Zain explains it well
ok
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