## And That Is The Rest Of The Story

Ask me anything   throw idea's my way!   Hello everyone, my name is Aidan Patterson, and I am the owner of my own thoughts and actions. My hobbies include Mathematics, Karate, Physics and anything music related. Never be afraid to tell me I am wrong, but be prepared to prove it, and may I say; good luck proving it to me!

Proofsareart

I’ve you’ve never been on proofsareart’s blog, you have not lived to see amazing mathematics!

— 1 week ago with 17 notes
#mathematics  #maths  #mathema  #physics  #math
Reblog or like if you have synesthesia!

That way, you can meet other synesthetes by looking at the notes :)

Numbers to colors, letters to colors, music to colors, instrument timbers to colors and textures, weekdays and other nonphysical objects including equations to spatial projections and personalities to colors. I’m still not sure how that last one works though :P

(via emsloe)

— 1 week ago with 174 notes
#Synesthesia

I hate when people are like: “You’re gay? So that’s why people like you!”. Like, no, people don’t like me more than you because I’m gay, I have several other personality traits that make me better than you.

— 1 week ago with 35 notes
#gay  #anger  #funny  #this happened today  #people  #suck  #dicks

"The Rape Poem to End All Rape Poems."

One of the best pieces of group spoken word poetry I’ve ever seen. WATCH IT.

THIS NEEDS TO BE WATCHED CHILLS FIRING THE WHOLE TIME POWER TO THE PEOPLE OF GOOD INTENTION AND LOVE

i started crying

this is what I mean when I say that anger is a powerful tool

— 1 week ago with 207355 notes

The proof of Sum of Square Numbers!

This is so much cooler than the proof for the Pythagorean theorem :)

— 3 weeks ago with 2308 notes
Pi is Irrational

So, a few days ago I found a proof for the irrationality of πand I thought I would share it with you all. The proof was originally done by Adrien-Marie Legendre in the 7th edition of his book elements de geometrie (1808). The version of this proof that I’ll be posting is more modern, and was completed in 1946 by Carl Siegel and published in his book: Transcendental Numbers. In case you don’t want to read it or the math gets too harry to read (I’ve been there with this proof), the proof is one by contradiction, where if we assume that π2 is a rational number and sub it in rational form into an equation with integer coefficients, we would expect an integer value to come out, but instead get  number between 0 and 1 which is obviously not an integer.This proof is pretty long, so here we go!

Assume

$e^x \approx -\frac{A(x)}{B(x)}$

Where A(x) and B(x) are polynomials of degree n, such that for x≈0 the ratio becomes close to exact.

Now, assume we can also write this equation

$R(x)=B(x)e^x+A(x)|x\rightarrow0,R(x)\rightarrow 0$

Since both A(x) and B(x) are of degree n, we can write:

$A(x)=a_0+a_1x+a_2x^2+...+a_nx^n$

$B(x)=b_0+b_1x+b_2x^2+...+b_nx^n$

By the direct multiplication of B(x) with the power series of ex we get

$R(x)=(a_0+b_0)+(a_1+b_0+b_1)x+(a_2+b_1+\frac{1}{2!}b_0+b_2)x^2+...$

Suppose we want R(x) to start with the 2n+1 term, that is let us assume the first 2n+1 coefficients of R(x) vanish. This will allow R(x) to vanish as fast as x2n+1, which is faster than x itself.

We now have the three equations

$(a_0+b_0)=0$

$(a_1+b_0+b_1)=0$

$(a_2+b_1+\frac{1}{2}b_0+b_2)=0$

Notice that we have 2n+1 equations, but 2n+2 coefficients so despite the restriction of R(x) we can still have nontrivial solutions. Now, we begin our exploration of differential operators. We let

$\mathbf{D}^n\phi(x)=\frac{d^n}{dx^n}\phi(x)$

$\mathbf{D}^{-1}\phi(x)=\int_{0}^{x}\phi(t)dt$

Observe that the condition for R(x) to approach zero when x approaches zero is vital in that

$\mathbf{D}^{-1}\mathbf{D}\phi(x)=\int_{0}^{x}\frac{d}{dt}\phi(t)dt=\phi(x)-\phi(0)$

$\mathbf{D}\mathbf{D}^{-1}\phi(x)=\frac{d}{dx}\int_{0}^{x}\phi(t)dt=\phi(x)$

So taking R(x) to zero for R(0) will allow the operators to commute. We now have to find a way to extend the definition for the inverse operator for all n as opposed to the singular definition we have now. Just to get a feel for the inverse operator for other values of n, let’s see how the n=2 case looks.

$\mathbf{D}^{-2}\phi(x)=\int_{0}^{x}\left ( \int_{0}^{s}\phi(t)dt \right )ds$

This integral has a nice geometric interpenetration. With t on the y-axis and s on the x-axis, the region we are integrating is under the line s=t, meaning we can make the substitution:

$\int_{0}^{s}\phi(t)dt=\int_{t}^{x}\phi(s)dt$

Which changes the integral to:

$\int_{0}^{x}\phi(t)\left ( \int_{t}^{x}ds \right )dt=\int_{0}^{x}\phi(t)(x-t)dt$

Which is again one dimensional. You can do the induction proof on your own, but this is already going to be too long, so I’ll just give the generalization for all n.

$\mathbf{D}^{-(n+1)}\phi(x)=\int_{0}^{x}\phi(s)\frac{(x-s)^n}{n!}ds$

We can use this to investigate integrals we’ll come across later. Now, we must investigate functions in the form:

$f(x)=e^{\lambda x}P(x)|x\rightarrow 0,\, P(x)\rightarrow 0$

We can see that taking the first derivative gives

$\mathbf{D}e^{\lambda x}P(x)=\lambda e^{\lambda x}P(x)+e^{\lambda x}\frac{dP}{dx}$

Taking the second derivative begins to show a pattern for nth derivatives, as:

$\mathbf{D}^2e^{\lambda x}P(x)=\mathbf{D}^1\left [ \lambda e^{\lambda x}P(x)+e^{\lambda x}\frac{dP}{dx} \right ]$

$\mathbf{D}^1\left [ \lambda e^{\lambda x}P(x)+e^{\lambda x}\frac{dP}{dx} \right ]=\mathbf{D}^1e^{\lambda x}\left [ \lambda P(x)+\frac{dP}{dx} \right ]$

$=\lambda e^{\lambda x}\left [ \lambda P(x)+\frac{dP}{dx} \right ]+e^{\lambda x}\left [ \lambda \frac{dP}{dx}+\frac{d^2P}{dx^2} \right ]$

$=e^{\lambda x}\left [ \lambda^2 P(x)+2\lambda\frac{dP}{dx}+\frac{d^2P}{dx^2} \right ]$

$=e^{\lambda x}\left [ \lambda^2 P(x)+2\lambda\mathbf{D}P(x)+\mathbf{D}^2P(x) \right ]$

$=e^{\lambda x}(\lambda + \mathbf{D})^2P(x)$

Again, I’ll skip the induction and just say that

$\mathbf{D}^ne^{\lambda x}P(x)=e^{\lambda x}(\lambda + \mathbf{D})^nP(x)$

Back to our formula for R(x),

$\mathbf{D}^{n+1}R(x)=\mathbf{D}^{n+1}e^xB(x)$

We can say this because A(x) was assumed to be degree n, and therefore disappears when differentiated n+1 times. From before, we can get a formula for the term containing B(x) by using the formula we just derived and setting λ=1.

$\mathbf{D}^{n+1}R(x)=e^x(1+D)^{n+1}B(x)$

Which means we can immediately write

$e^{-x}\mathbf{D}^{n+1}R(x)=(1+D)^{n+1}B(x)$

Since R(x) starts with the 2n+1 term, we can say that:

$\mathbf{D}^{n+1}R(x)=(2n+1)(2n)(2n-1)...(n+1)r_1x^n$

Plus all of the other terms of degree n+1 and higher, where ris the first coefficient of R(x). Because B(x) was assumed to be a polynomial of degree n,

$(1+\mathbf{D})^{n+1}B(x)$

Will have to be degree n as well, as one term will contain 1B(x), and the others will all contain some constant times some derivative of B(x) which will be a lower degree polynomial. If we Say that

$r_0=(2n+1)(2n)(2n-1)...(n+1)r_1$

Then we may only keep the leading term of R(x) in our expansion. If we let e^x take on its power expansion again, we see the only term that need be considered again is $r_0x^n$.So,

$B(x)=(1+\mathbf{D})^{-(n+1)}r_0 x^n$

We can use a similar method to derive A(x) (multiply through the original expression for R(x) by e^(-x) and then do the work with derivatives we did here) which works out to:

$A(x)=(-1+\mathbf{D})^{-(n+1)}r_0 x^n$

Now, recall that

$\mathbf{D}^{n+1}R(x)=e^x(1+\mathbf{D})^{n+1}B(x)$

Using our solution for B(x) that we just derived, we can see that

$\mathbf{D}^{n+1}R(x)=e^xx^n$

Or,

$R(x)=\mathbf{D}^{-(n+1)}e^xx^n$

From before

$\mathbf{D}^{-(n+1)}\phi(x)=\int_{0}^{x}\phi(s)\frac{(x-s)^n}{n!}ds$

So,

$R(x)=\frac{1}{n!}\int_{0}^{x}s^n(x-s)^ne^{s}ds$

Letting u=(s/x), du=1/x, we see that

$R(x)=\frac{x^{2n+1}}{n!}\int_{0}^{1}u^n(1-u)^ne^{ux}du$

If we let t=1-u and dt=-du then,

$R(x)=\frac{x^{2n+1}}{n!}\int_{0}^{1}t^n(1-t)^ne^{(1-t)x}dt$

If we change the dummy variable of integration in the line

$R(x)=\frac{x^{2n+1}}{n!}\int_{0}^{1}u^n(1-u)^ne^{ux}du$

To t, add it to the previous expression and then divide by 2, we see that:

$R(x)=\frac{x^{2n+1}}{n!}\int_{0}^{1}t^n(1-t)^n\frac{e^{tx}+e^{(1-t)x}}{2}dt$

Changing the form a bit, we get

$R(x)=\frac{x^{2n+1}}{n!}e^{x/2}\int_{0}^{1}t^n(1-t)^n\frac{e^{(t-1/2)x}+e^{-(t-1/2)x}}{2}dt$

We can now sub in a value for x to make further simplifications on this integral, let x=iπ.

$R(i\pi)=\frac{(i\pi)^{2n+1}}{n!}e^{i\pi/2}\int_{0}^{1}t^n(1-t)^n\frac{e^{(t-1/2)i\pi}+e^{-(t-1/2)i\pi}}{2}dt$

Some quick simplification yields:

$R(i\pi)=(-1)^{n+1}\frac{\pi^{2n+1}}{n!}\int_{0}^{1}t^n(1-t)^nSin(\pi t)dt$

Now, we’re not evaluating this. I just want you to note that depending on the parity of n, the expression for $R(i\pi)$, we can get either a positive or negative number, but not zero. Therefore, $R(i\pi)\neq 0$. Now, we have to do a bit of manipulation with the equation that we had for R(x) in the beginning.

$R(x)=e^xB(x)+A(x)$

Immediately it is clear that

$R(-x)=e^{-x}B(-x)+A(-x)$

and

$e^xR(-x)=B(-x)+e^xA(-x)$

Remember, our original first alternate expression for R(x),

$R(x)=\frac{1}{n!}\int_{0}^{x}(x-s)^ne^ss^nds$

Altering it to fit the previous expression gives

$e^xR(-x)=\frac{e^x}{n!}\int_{0}^{-x}(-x-s)^ne^ss^nds$

For t=-s and ds=-dt,

$e^xR(-x)=\frac{-1}{n!}\int_{0}^{x}(x-t)^ne^{x-t}t^ndt$

again for u=x-t, du=-dt

$e^xR(-x)=\frac{-1}{n!}\int_{0}^{x}(x-u)^ne^{u}u^ndt$

Which makes it clear that

$e^xR(-x)=-R(x)$

Now we can get two sets of equations for A(x) and B(x)

$R(x)=e^xB(x)+A(x)$

$R(x)=-B(-x)-e^xA(-x)$

Subtracting the second from the first give

$R(x)=e^x[B(x)+A(-x)]+[A(x)+B(-x)]$

This can be true for all x if and only if the brackets vanish identically, that is:

$B(x)=-A(-x)$

$B(-x)=-A(x)$

These are equivalent statements, and they can be brought together by again setting $x=i\pi$.

$B(i\pi)e^{i\pi}+A(i\pi)=R(i\pi)$

So,

$A(-i\pi)+A(i\pi)\neq0$

as

$R(i\pi)\neq 0$

Since A(x) is a polynomial in x, we can expand it from here and see where this fact leaves us.

$A(x)=a_0+a_1x+a_2x^2+...+a_nx^n$

$A(-x)=a_0-a_1x+a_2x^2-...\pm a_nx^n$

Where the final term in A(-x) depends on the parity of n. Adding these expressions together as before gives:

$A(x)+A(-x)=2a_0+2a_2x^2+2a_4x^4+...$

where the final term is either of degree n or n-1. If we let [m] be the integer part of a number m, and change the variable from x^2 to u, we can see:

$A(x)+A(-x)=2a_0+2a_2u+2a_4u^2+...+ 2a_{2[n/2]}u^{[n/2]}$

Now, here comes the end, suppose that π2 is rational, then with x=iπ and u=-π, it follows that

$A(x)+A(-x)=2a_0-2a_2\frac{p}{q}+2a_4\frac{p^2}{q^2}-...\pm 2a_{2[n/2]}\frac{p^{[n/2]}}{q^{[n/2]}}$

Since all of the coefficients a_i were assumed to be integers, then

$q^{[n/2]}R(i\pi)=j$

where j is some integer, positive or negative. Since we know that

$R(i\pi)\neq 0$

then,

$|q^{[n/2]}R(i\pi)|=|j|>0$

Finally, from before,

$|R(i\pi)|=\left | (-1)^{n+1}\frac{\pi^{2n+1}}{n!}\int_{0}^{1}t^n(1-t)^nSin(\pi t)dt \right |$

Taking limits, we can see that both the coefficient and the integral approach zero as n approaches infinity, so we can always chose sufficient n such that

$|q^{[n/2]}R(i\pi)|<1$

Which of course implies

$0<|q^{[n/2]}R(i\pi)|=|j|<1$

So,

$0<|j|<1$

This means that there should be an integer between zero and one, which is an obvious contradiction. This is sufficient to show that π2  is irrational, and it follows that π is irrational

$\square$

Have a happy pi month :)

— 3 weeks ago with 28 notes
#math  #mathematics  #maths  #mathema  #physics  #integrals  #inequalities  #pi  #e  #euler  #irrational  #number theory  #calculus  #operators  #proof
Transcendental Pi

I’m going to get a proof that pi squared is irrational (and therefore pi is irrational) on here soon, but in the meantime, have a simple proof that pi is transcendental :)

The Lindermann-Weierstrass theorem states that for the equation

$e^a=b$

b will always be transcendental if a is a non-zero algebraic number.

From this, we make note that

$e^{i\pi}=-1$

Meaning either i or pi have to be transcendental. Because i satisfies the relation:

$x^2+1=0$

Pi must be transcendental.

The real gem of this proof is the Lindermann-Weierstrass theorem, which I will look into finding a proof for soon. I’m about halfway through the proof that pi squared is irrational, so stay tuned, because it should be up by Friday :)

— 1 month ago with 31 notes
#math  #mathematics  #maths  #physics  #mathema  #pi  #i  #theorem  #transcendental  #numbers  #number theory
Quantum states and superposition →

Sit down, kids, you’re about to learn a thing.

(This will have significant overlaps with the Quantum Physics Sequence, albeit with more mathematics and symbols and what I hope will be a different angle. I will also shamelessly steal pictures from there.)

Let’s start…

If you want to do these experiments, the new 3-D glasses they give out at movie theaters have polarized lenses in them. You could just pop them out and start doing science :D

— 1 month ago with 23 notes

My math binders are always red every year I feel like math is just a red subject

Math is a blue subject and I’m prepared to fight you over this

OKAY NO! Math is red. Red. RED. RED!!! Why?

Capital M’s have a deep red color to them, (The M in Monarch feels the exact same way)

the “ath” part of Math is actually a light blue color (kind of like a summer sky, the a is white, the t is blue, and the h is an indigo), which is where you might be getting the blue from, but with a “M” at the beginning, the word is overpowered by the red hue that changes the a from white to a pink, making the word feel more red with the addition of the h.

I wonder if the difference here is the part of the word people focus on when they’re reading?

Mathematics on the other hand is a very intricate word.

the “M” at the beginning is a strong red again, but the e after “ath” is black, which makes the second m seem bright yellow. The “at” is white again, but the “ic” is strongly red, and the s is bright yellow.

This doesn’t mean that the overall feel of the word is orange though, and I actually feel that the word is distinctly red in the “Math” section, black on the “e” and finishes yellow with “matics”. Shoved together, I get a pinkish feeling, but I can very clearly see yellow and black.

And yes, my math binders are always red or pink.

On a side note, this is reason I hate the word maths, because in the presence of a yellow s, the red math turns to pink, and makes the word look like an Easter egg, which I don’t feel is very representative of mathematics.

I love synesthesia  :)

— 1 month ago with 107742 notes
#math  #mathematics  #maths  #physics
Cauchy-Schwarz Inequality, Time, and Space

Say you’re sitting at home, and you thought to yourself, if I threw myself off the 40th floor balcony in my apartment building, what would my average speed be when I hit the ground? You don’t really intend to jump to find out, so you start doing some math. At this point, 9,999 out of 10,000 people would say that since the acceleration of gravity is 9.8 meters per second per second, and assuming each floor was about ten feet long (total of 122 meters), you would probably guess that your average speed would be about 24.5 m/s and leave it at that. BUT you are a mathematician (which is probably why you were contemplating jumping off a building in the first place), and you want to make sure you’re right. What you just calculated was a time average, that is:

$V_{time}=\frac{1}{T}\int_{0}^{T}v(t)dt$

But, what if you wanted a different viewpoint? What if you wanted to look at the speed as a function of the distance fallen just to make sure the number you got was the right one? The equation is remarkably similar,

$V_{space}=\frac{1}{L}\int_{0}^{L}v(y)dy$

but, gives you strange answers. Since L=122 and y=4.9t^2, we can see that:

$V_{space}=\frac{1}{122}\int_{0}^{122}9.8(y/4.9)^{\frac{1}{2}}dy$

$V_{space}=\frac{\sqrt{19.6}}{122}\int_{0}^{122}\sqrt{y}\, \, dy$

$V_{space}=\frac{\sqrt{19.6}}{122}\left ( \frac{2}{3}(y^{3/2})\right |^{122}_0$

$V_{space}=\frac{\sqrt{78.4}}{366} (122)^{3/2}$

$V_{space}\approx 32.6 \, m/s$

That’s obviously quite different from 24.5 m/s. You could argue (probably for hours) about the meaning of this result, but since you’re a mathematician and not a physicist, you’re more interested in whether

$V_{space}\geq V_{time}$

regardless of the boundary conditions. You need a way to compare the two, so you think to yourself, what kind of math could I do with two integrated functions?

$\int_{U}^{L}(f(t)+\lambda g(t))^2dt\geq 0$

If f(t) and g(t) are real valued functions, and U, and, L are real numbers (possibly infinite) and, lambda is real (for now).

Expanding, we can see that:

$\int_{U}^{L}f^2(t)dt+2\lambda\int_{U}^{L}f(t)g(t)dt+\lambda^2\int_{U}^{L}g^2(t)dt\geq 0$

If we label each separate definite integral c,b,a respectively, we get:

$a\lambda^2+2b\lambda+c\geq 0$

This has a simple geometric interpretation, the function of lambda can not cross the lambda-axis, the most it can do is touch it, as we are dealing with the area under functions. This must mean that lambda is complex for h(lambda)=0, or has a double zero at that point.

$\lambda=\frac{-b\pm \sqrt{b^2-ac}}{a}$

So, we get the inequality:

$b^2\leq ac$

This finally gives the inequality

$\left ( \int_{U}^{L}f(t)g(t)dt \right )^2\leq \left ( \int_{U}^{L}f^2(t)dt \right )\left ( \int_{U}^{L}g^2(t)dt \right )$

So how are we going to use this to see if your velocity measured with time is less than or equal to your velocity measured using space?

Well, we can let g(t)=v(t)/T, and let f(t)=1. The Cauchy-Schwarz Inequality says that:

$\left ( \frac{1}{T}\int_{0}^{T}v(t)dt \right )^2\leq \left ( \int_{0}^{T}dt \right )\left ( \frac{1}{T^2} \int_{0}^{T}v^2(t)dt \right )$

and

$\left ( \int_{0}^{T}dt \right )\left ( \frac{1}{T^2} \int_{0}^{T}v^2(t)dt \right )=\frac{1}{T}\int_{0}^{T}v^2(t)dt=\frac{1}{T}\int_{0}^{T}\left ( \frac{dy}{dt}\right )^2dt$

$\left ( \frac{dy}{dt} \right )^2dt= \frac{dy}{dt}\cdot \frac{dy}{dt}\cdot dt=\frac{dy}{dt}dy$

We’ve now effectively changed the variable to which we are integrating, so we have to change the limits on the integral, namely T=L and t=y

So,

$\frac{1}{T}\int_{0}^{T}v^2(t)dt=\frac{1}{T}\int_{0}^{L}v(y)dy$

We’re really close now! Since

$L=\int_{0}^{T}v(t)dt$

We can change the left side of the inequality to be

$\left ( \frac{1}{T}\int_{0}^{T}v(t)dt \right )^2=\frac{L}{T^2}\int_{0}^{T}v(t)dt$

Which gives us

$\frac{L}{T^2}\int_{0}^{T}v(t)dt\leq \frac{1}{T}\int_{0}^{L}v(y)dy$

and finally,

$\frac{1}{T}\int_{0}^{T}v(t)dt\leq \frac{1}{L}\int_{0}^{L}v(y)dy$

So,

$V_{time}\leq V_{space}$

Strange isn’t it?

I’ve deliberately left a bunch of questions unanswered, in the hopes that you’ll look into them. For one, what is the physical interpretation of the velocity in space that we derived in the first section? Also, would drag affect the answer in any way?

In any case, I hope this at least made you think. Pretty cool right? :)

— 1 month ago with 74 notes
#mathematics  #maths  #math  #mathema  #physics  #integrals  #complex  #numbers  #velocity  #time  #space  #relativity  #calculus  #analysis  #dynamics  #proof
nelolin asked: Hi there, I searched 'gamma function' on tumblr and found your post about fractional derivatives. I am glad I found your blog. :). On another note, do you use a Latex extension for your posts? Or MathJax?

Hey! sorry about talking forever to respond, I have a bunch of messages and tumblr does a terrible job of telling me when I have one -.- I use LaTeX for my posts, but more specifically, I use this website: http://www.codecogs.com/latex/eqneditor.php and then just copy the images that it spits out :)

— 1 month ago with 6 notes

Brussel  Sprouts - Numberphile

Some brussel sprouts are good for you, others will teach you about topology and the Euler characteristic.

— 1 month ago with 32 notes

Solutions to the general equations

ax+b=0

ax^2+bx+c=0

ax^3+bx^2+cx+d=0

ax^4+bx^3+cx^2+dx+e=0

There does not exist a general solution by radicals for equations of degree 5 or more - a fact that can be proved using Galois Theory and the fact that Sn is not solvable for n greater than or equal to 5.

— 1 month ago with 475 notes

Honey, I don’t think you gotta worry about that anytime soon.

A few things.

1) Your TAGS are DISGUSTING. You are a misogynistic, fat-shaming, rape apologist. The idea that a woman, regardless of how she looks, would be OPTIMISTIC to be raped is just stomach churning.

2) The image you have posted is CLEARLY PHOTOSHOPPED. Here is this woman’s original picture with her ACTUAL message:

Thanks for proving her point.

You gross ass sexist crumblefuck.

I didn’t even make the picture, I fucking found it on the internet. Thanks for not being able to take a fucking joke.

Oh wow here we fucking go again, but let’s go in categories this time to make this easy for you.

Rape

There is so much more than can be said but fuck you, go look that shit up yourself.

Fat Shaming

• Fat shaming is damaging, bullying and wrong period. It is fucked up to make fun of someone for any reason. The only reason why someone should be spoken to in a negative fashion is if they have actually done something wrong and harmful. (Like you did!)
• The BMI scale is a crock of shit. It doesn’t actually determine what you as an individual need to be functioning properly. It goes by a shitty, rigid “weight by height” requirement. It doesn’t take into account personal family genetics, for instance.
• Some people are just fat. That’s it. Genetics, disorders, pregnancy, diseases, illness and body composition just fucking happen to some people. Body shaming does not make fat go away.
• White culture leads to rigid beauty standards: thin, white, able-bodied and minded, symmetrical face with fat in all the “right places”. Only about 10% of women actually fit this “standard” and are ultimately starving for it.
• Fat people suffer from not only peer abuse and media ridicule, but eating disorders and other heinous things to try and match their thinner counterparts.
• Fat people are systematically oppressed. IE they don’t get jobs like their thin counterparts based on people thinking they are “lazy” and other things.
• A case in Australia about how a couple could not adopt a child because the woman was too fat. (Also, England.)
• None of this shit actually has to do with these people as individuals, their skill sets, etc. This is discrimination against them as being fat only, regardless if they can “help it” or not. (IE genetics versus simply lifestyle.)

So fuck you.

THANKS FOR THE SOURCES! I AM USING THEM IN AN ENGLISH ESSAY!

(via kougannohime-chan)

— 1 month ago with 3282 notes

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