shah_nosrat

There's a mistake in your equality; it's supposed to be [latex]2xy^2 - y^2 + y = 2yx^2 - x^2 + x [/latex], factoring both terms on each side of the equality we have: [latex] y(2xy -y + 1) = x(2xy -x + 1) [/latex], which we can argue for x to equal to y we should have [latex]2xy -y + 1 = 2xy - x + 1[/latex], which reduces to [latex]x=y[/latex] as required, hence, your function is injective.

From what I studied in discrete mathematics, relations have matrix representations, and besides the operations we are used in linear algebra, there are also boolean operations that can be performed on matrices.

I don't understand your vector representations. But from the definition of linear independence, it says the following, that the linear combination of the vectors; k_{1}v_{1} + k_{2}v_{2} + k_{3}w = 0, if k_{1}=k_{2}=k_{3}=0 solving this equation will give your solution.

Just to answer your last statement. Abstract Algebra deals with the study of mathematical structures called groups. To give an example between LA (Linear Algebra) and AL (Abstract Algebra) on their similarity (not in property, but on the approach, since it's algebra). In LA we have basis sets that spans a particular vector space, and how an entire vector space can be constructed by the basis set. Similarly in AL we have cyclic sets that generates an entire group! So their approach is similar, but as John puts it: that's why is called Linear Algebra. Besides, LA has immense applications; handwriting analysis, solving linear first-order differential equations, search engines use the mathematics from LA, Differential Geometry: representation of the coefficients of the first-fundamental form is in matrix form. Hope this clarifies things.

I haven't read the above book. But I do know that any computer scientist needs to have knowledge of Discrete Mathematics: As this will teach you naive set theory, logic, counting principles, Relations, Digraphs, Graph theory, Languages and finite - state machines and much more. Then you could complement it with the above mentioned book. When dealing with Definitions, axioms, theorems, and proofs. It's always a good idea to understand what a particular definition, axiom or theorem is saying and then going on to reading the associated proof to get a complete picture of whats going on. Memorizing is never a good idea.

Edward Witten Michio Kaku Andrew Wiles Stephen Hawking Roger Penrose Yes some of the above are Theoretical Physicists, but as ajb put it, they did spur modern mathematics.

I don't know if considering robots as its own species would be a good idea, not to mention it being an intellectual being/entity. We would then have to consider their robotic rights as well, and would give rise and debate to ethical considerations of how to deal with these new robotic beings or their species as a whole. But I do know that Japanese scientists creating realistic humanoid robots to assist us in our daily chores or life for that matter, they consider it as being the next evolution of the so-called "Personal Computer". Now to consider something as alive, they would have to satisfy certain prerequisites (Which I'm not really sure of) but I'm sure it exists.

Hi, try some books by Stephen Hawking: George's Secret Key to The Universe, and other science books for Kids. I don't know if they improve critical thinking, but it sure will encourage imagination and creative thinking in kids, which is also important in any field of science.

Hi, I came across this theorem and decided to prove it, as follows: Theorem: A set [math]A \subseteq R[/math] is bounded if and only if it is bounded from above and below. I would like the prove the converse of the above statement; If a set [math]A \subseteq R[/math]is bounded from above and below, then it is bounded. Let [math]M = |M_{1}| + |M_{2}|[/math]and using this preliminary result I proved earlier [math]-|a| \leq a \leq |a|[/math]. Now, [math]\forall a \in A[/math] we have [math] a \leq M_{1}[/math] ---> definition of bounded from above. and [math] M_{2} \leq a[/math] ---> definition of bounded from below. Using the result: [math]-|a| \leq a \leq |a|[/math]. Since [math]M = |M_{1}| + |M_{2}|[/math] is the sum of absolute value of [math]|M_{1}|[/math] and [math]|M_{2}|[/math], it is a big number. (trying to convince myself). Also, [math]-M = -(|M_{1}| + |M_{2}|)[/math], This is on the opposite of the spectrum. Now, [math]-M \leq -|M_{2}| \leq M_{2} \leq -|a| \leq a \leq |a| \leq M_{1} \leq |M_{1}| \leq M [/math]. [math]M = |M_{1}| + |M_{2}| \geq |M_{1} + M_{2}| \geq 0 [/math] ---> Which shows that M is positive by using the triangle inequality. Hence, [math]-M \leq a \leq M[/math]. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% I'm excited for this proof, hopefully it's correct. Your help is once again appreciated.

Thank you Dr.Rocket. You are giving me confidence in art of proofing. Thank you Dr.Rocket. You are giving me confidence in art of proofing.

Okay, taking your advice. Forget about my previous attempt at the proof. Using Linear (total) ordering. If we take [math]\mathbf{U} \subseteq W[/math] ---> I need to invoke existence for it to make sense (To me anyway) suppose [math]a, b \in \mathbf{U}[/math] with the property of being the least elements for all elements in [math]\mathbf{U}[/math]. Now, because of the antisymmetry property of the linear ordering then, as follows: if aRb and bRa then a = b ---> Does this conclude the least element in the subset if it exist is unique? Your help is once again appreciated

Seattle's Calling - by Burn The Charts

Hi, This is the question that needs a proofs, as follows: Show that the smallest element of a nonempty subset of [math]\mathbf{W}[/math] is unique. My attempt at the proof, as follows: Let [math]\mathbf{U} \subseteq \mathbf{W}[/math], by the well ordering principle (WOP) we have that [math]a \in \mathbf{U}[/math] such that [math]a \leq x [/math] [math]\forall x \in \mathbf{U}[/math]. Now suppose [math]b \in \mathbf{U}[/math] such that [math]b \leq x [/math] [math]\forall x \in \mathbf{U}[/math]. Since [math]0 \leq x - a[/math] and [math]0 \leq x - b[/math] by definition. Now, [math]0 \leq x + x - (a + b)[/math] [math]a+ b \leq x + x = 1\cdot x + 1\cdot x = (1+ 1)\cdot x = 2x[/math] [math]a+ b \leq 2x[/math] [math]\frac{a + b}{2} \leq x[/math] . The only way that this inequality will hold [math] \forall x \in \mathbf{U}[/math] is when [math]a = b[/math] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Is the above reasoning and proof correct?

First of all keep the doubts and negative thinking aside. I believe that anyone is capable of great things as long as they're interested. If you are interested in Higher Mathematics then that's the first step to learning Higher Mathematics. Let me tell you this, that Higher Mathematics trains you to think analytically and critically about any problem presented to you. And remember to always have fun