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I already showed you why your understanding of omnipotence is silly and incoherent. I also showed how the statements "God has power over all things" and "God's power is unlimited" are also coherent, since these impossibilities, such as a square-circle or a married-bachelor are merely ideas and...

It is the view among most philosophers as well, and in fact philosophers would be more likely to hold my view of omnipotence since some theologians can in fact be anti-rationalist and would take the position you take. I don't know why you are pressing this point, it is quite clear that no one...

That is the layman's definition, it is the consensus of all theologians, across pretty much every religion, with only very few exceptions, that omnipotence only refers to that which is possible. If you want to criticise those people (like Decartes iirc) who believed that God can perform the...

Re: HSC 2015 4U Marathon \\ $Let there be a triangle$ \ ABC \ $and points$ \ P,Q,R \ $on$ \ BC, AC, AB \ $respectively, such that$ \ P,Q,R \ $are midpoints on the sides$ \ BC,AC,AB \ $respectively$ \\ \\ $Show that$ \\ 4(AP^2 + BQ^2 + CR^2) = 3(AB^2 + AC^2 + BC^2)

Re: HSC 2015 4U Marathon - Advanced Level \\ $Let$ \ a_2, a_3, \dots , a_n \ $be among the positive reals so that$ \ a_1 a_2 a_3 \dots a_n = 1 \\ \\ $Prove that$ \ (1+a_2)^2(1+a_3)^3(1+a_4)^4\dots (1+a_n)^n > n^n

No theologian in their right mind would say that God can "defy" logic. Not even God can lift a stone that he cannot life, God cannot make 1 + 1 =3, God cannot make that which is round, square, God cannot make a ball that is both in motion and in rest at the same time and in the same respect. An...

I know its not a proof, its a clarification of a definition. If someone shows that there is an uncaused being with a rational argument (who for all intents and purposes we will call God), then to then ask what caused God is simply irrational.

God does not have a cause since, God, being the most perfect of all beings is uncaused and Necessary in existence. Therefore, an argument for God's existence is to argue for an uncaused being, to ask then "what caused this uncaused being" doesn't make sense, misses the point of the argument and...

The Universe cannot be the ultimate existent as it is a contingent existent, it is a contingent existent because the laws of that govern the Universe and the way the Universe operates is in no way true by necessity. If the Universe was necessary then everything about the Universe would be...

Well it comes directly from the definition Two conjugates of each other are defined as: z = r(\cos \theta + i\sin \theta) \overline{z} = r(\cos (-\theta) i\sin (-\theta)) z \cdot \overline{z} = r^2 (\cos(\theta -\theta) + i\sin(\theta - \theta)) = r^2 So, if I have 2 complex numbers of...

So remember that w is a root of z^5 = 1 This means that w^5 = 1, right? Since w^5 = 1 and w^4*w = w^5 (since when we multiply them the powers add) thus w^4 * w = 1 So what we have here is 2 complex numbers, that multiplied together get a real number (and both complex numbers have the same...

I am willing to bet that there is a typo in what \alpha is supposed to be \alpha = w + w^2 makes more sense with part (iv) Anyway, for a complex number z, its conjugate is such that z*conj(z) = real number So, since w \cdot w^4 = w^5 = 1 thus w and w^4 are conjugates

I have a feeling that there is a typo in the question and that \alpha = w + w^2 Because as it is being asked now, the answers are trivial, and they don't even aid in getting part (iv)

Re: HSC 2015 4U Marathon \\ 8x^3 - 3x^2 - 3x - 1 = 0 \\ 8x^3 = 3x^2 + 3x + 1 \\ 9x^3 = x^3 + 3x^2 + 3x + 1 \\ \Rightarrow \ 9x^3 = (x+1)^3 \\ \\ \therefore \ \left(\frac{x+1}{x} \right)^3 = 9 \\ $So,$ \ \frac{x+1}{x} = \sqrt[3]{9} \Rightarrow \ x = \frac{1}{\sqrt[3]{9} - 1} =...

Re: HSC 2015 4U Marathon Typo the first x^3 is supposed to be x^4, fixed

Re: HSC 2015 4U Marathon \\ x +x^3 - x^4 - x^6 = (x-x^4) + (x^3-x^6) = x(1-x^3) + x^3(1-x^3) = (x+x^3)(1-x^3) \\ $For$ \ x \geq 1 \ $and for$ \ x \leq 0 \ $then$ \ (x+x^3)(1-x^3) \ $is negative, so is less than 1$ \\ \\ $For$ \ 0 < x < 1 \ $then both terms$ \ (x+x^3) \ $and$ \ (1-x^3) \ $are...

Re: HSC 2015 4U Marathon \\ $The real root of the equation$ \ 8x^3 - 3x^2 - 3x - 1 = 0 \ $has a real root of the form$ \ \frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c} \ $where$ \ a,b,c \ $are positive integers$ \\ \\ $Find$ \ a + b + c

Re: HSC 2015 4U Marathon \\ $Prove the inequality$ \ x + x^3 - x^4 - x^6 < 1 \ $for all real$ \ x

Re: HSC 2015 4U Marathon - Advanced Level \\ $Define the sequence$ \ x_1 = \frac{1}{2} \ , \ x_n = \frac{x_{n-1}}{2nx_{n-1} + 1} \\ $Find$ \ \sum_{k=1}^n x_k

Re: HSC 2015 4U Marathon - Advanced Level My method: \\ \sum_{k=2}^{n+1} \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}} < \frac{1}{2\sqrt{2}}t_n < \sum_{k=1}^{n}\frac{1}{\sqrt{2k-1} + \sqrt{2k+1}} \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}} = \frac{1}{2} (\sqrt{2k+1} - \sqrt{2k-1}) \Rightarrow \...