Is (x,y)<(j,k) Defined as an Order Relation by x+k<y+j?

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Homework Statement



Q. Define an order relation
(x,y)<(j,k) if and only if x+k<y+j

Homework Equations


x<y means x+a=y and viceversa

The Attempt at a Solution


I have no idea. To show it is equivalence relation, I simply show that it is reflexive, transitive and symmetric. but how do I show it is order relation?
 
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Order Relation (call it C)
1) For every x and y, with x not equal to y, xCy or yCx
2) xCx is never true
3) xCy, yCz implies xCz
 
dpa said:

Homework Statement



Q. Define an order relation
(x,y)<(j,k) if and only if x+k<y+j


Homework Equations


x<y means x+a=y and viceversa
With a> 0, of course.


The Attempt at a Solution


I have no idea. To show it is equivalence relation, I simply show that it is reflexive, transitive and symmetric. but how do I show it is order relation?
An order relation only has to be transitive.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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