Your device is currently offline. You can view downloaded files in for Languages & International Education (CLIE) offers a range of 0.5 and 1.0 course units (15 and 30 credits/7.5 and 15 ECTS respectively). Registration takes place from Friday 23 September to Friday 30 September 2016 (Monday to Friday 10:00-17:00, no appointment necessary, and you need to come to register in person). Classes will be timetabled between 03 and 05 October 2016 (you will be given the day/time for your module when you see a language tutor during registration), and classes start in the week of 10 October 2016. Modern foreign language course units English for Academic Purposes course units Teaching English as a Foreign Language course unit British Sign Language 0.5 course unitPlease check the answers to questions we often get asked. If your question is not answered then please email us. Much of this site is restricted to current UCL staff and students. For access you will require a valid UCL Information Services Division computer account and password.

We shall now discuss some basic theorems related to rings. We feel that a good way to learn ring theory is to try out proofs of simple theorems on ones own. Hence the reader is encouraged to work out proofs of theorems by him/herserlf before reading the proofs given here. Often we shall provide only a sketch of the proof and the reader is expected to fill in the gaps in that case. Theorem 1.1: If R is a ring and ; 1. a+b=a+c implies b=c. (Cancellation Law) 3. The zero element of R is unique. 4. The additive inverse of any element is unique. 1. Clearly adding -a on both sides of a+b=a+c gives us the desired result. 2. It suffices to show that a+(-a)=0 which is obvious from the definition of -a. 3. If there exists two zero elements 0 and 0' in R then 0+0'=0' and 0+0'=0 by definiton and so 0=0'. 4. If a' and a'' are two inverses of a then a'=a'+0=a'+a+a''=0+a''=a''. Theorem 1.2: If R is a ring, then for any ; If in addition, R has a unit element 1, then

By the cancellation law it now follows that a0=0. 2. It suffices to show that a(-b)=-(ab) or equivalently a(-b)+ab=0. Now a(-b)+ab=a(b-b)=a0=0 by 1. and so the result is proved.Again -(a(-b))=-(-(ab))=-(-ab) by 2. But by Theorem 1.1(2) -(-ab)=ab. 5. (-1)a+a=(-1)a+1a=(-1+1)a=0a=0 by 1. 6. Put a=-1 in 5. and apply Theorem 1.1(2). It is strongly recommended that theorems in this section should be treated as exercises by the readers. Theorem 1.3: Prove that a ring R is commutative if and only if holds for all . Proof: Suppose R is commutative. Then clearly the result holds. (In fact the binomial theorem: holds in that case. Try to prove it using induction and the Pascal's identity:.) Conversely suppose that for each the given relation is satisfied. Now, on applying the distributive laws to we get and by the cancellation laws we have ab=ba. Hence R is commutative. Theorem 1.4: If R is a system satisfying all the conditions of a ring with unit element with the possible exception of a+b=b+a, prove that the axiom a+b=b+a must hold in R and that thus R is a ring.

Proof: (a+b)(1+1)=a1+a1+b1+b1=a+a+b+b and (a+b)(1+1)=a1+b1+a1+b1=a+b+a+b by the left and right distributive laws reepectively. Equating the two identities and applying the cancellation laws gives us the result.cheap ac units in nyc Theorem 1.5: Let R be a ring such that for all . used wall ac unitProve that R is commutative.ac window unit for sale Note: Such a ring is called a Boolean ring.Since and so by the cancellation law. Now as so and so each element in R is its own additive inverse. Hence and so . Theorem 1.6: If R is a ring with unity satisfying for all , prove that R is commutative. Proof: By our hypothesis and also by the distributive laws . So equating the two and applying the cancellation laws we have which holds as an identity.

Now substituting x+1 for x in the identity we have . This gives and on the application of the distributive laws we have . Cancellation law now gives as required. Theorem 1.7: Let R be a ring such that for , there exists a unique such that xa=x. Show that ax=x. Hence deduce that if R has a unique right identity e, then e is the unity of R. Proof: xa=x implies x(a+ax-x)=xa+xax-x2=x. Hence a+ax-x=a or ax=x. If R has a unique right identity e then xe=x implies ex=x and so e is the unity of R. Theorem 1.8: Let R be a ring with unity . Suppose for a unique such that xyx=x. Prove that xy=yx=1, i.e. x is invertible in R. Proof: Suppose, if possible xa=0 for some . Now, x(y+a)x=(xy+xa)x=xyx+xax=xyx=x and by the uniqueness of y it follows that y+a=a i.e. a=0. Now x(yx-1)=xyx-x=x-x=0 and so yx-1=0. So x is invertible. Theorem 1.9: Show that if 1-ab is invertible in a ring R with unity, then so is 1-ba. Proof: Let x be the inverse of 1-ab, i.e. let x(1-ab)=(1-ab)x=1. So 1-ba is invertible with inverse 1+bxa.

Theorem 1.10: If a,b are any two elements of a ring R and m and n are any two positive integers, then prove that Proof We shall prove 4. and leave the rest as an exercise for the reader. 4. by repeated application of the distributive law. The RHS is just (nm)(ab).Let’s connect three AC voltage sources in series and use complex numbers to determine additive voltages. All the rules and laws learned in the study of DC circuits apply to AC circuits as well (Ohm’s Law, Kirchhoff’s Laws, network analysis methods), with the exception of power calculations (Joule’s Law). The only qualification is that all variables must be expressed in complex form, taking into account phase as well as magnitude, and all voltages and currents must be of the same frequency (in order that their phase relationships remain constant). KVL allows addition of complex voltages. The polarity marks for all three voltage sources are oriented in such a way that their stated voltages should add to make the total voltage across the load resistor.

Notice that although magnitude and phase angle is given for each AC voltage source, no frequency value is specified. If this is the case, it is assumed that all frequencies are equal, thus meeting our qualifications for applying DC rules to an AC circuit (all figures given in complex form, all of the same frequency). The setup of our equation to find total voltage appears as such: Graphically, the vectors add up as shown in Figure below. Graphic addition of vector voltages. The sum of these vectors will be a resultant vector originating at the starting point for the 22 volt vector (dot at upper-left of diagram) and terminating at the ending point for the 15 volt vector (arrow tip at the middle-right of the diagram): (Figure below) Resultant is equivalent to the vector sum of the three original voltages. In order to determine what the resultant vector’s magnitude and angle are without resorting to graphic images, we can convert each one of these polar-form complex numbers into rectangular form and add.

Remember, we’re adding these figures together because the polarity marks for the three voltage sources are oriented in an additive manner: In polar form, this equates to 36.8052 volts ∠ -20.5018o. What this means in real terms is that the voltage measured across these three voltage sources will be 36.8052 volts, lagging the 15 volt (0o phase reference) by 20.5018o. A voltmeter connected across these points in a real circuit would only indicate the polar magnitude of the voltage (36.8052 volts), not the angle. An oscilloscope could be used to display two voltage waveforms and thus provide a phase shift measurement, but not a voltmeter. The same principle holds true for AC ammeters: they indicate the polar magnitude of the current, not the phase angle. This is extremely important in relating calculated figures of voltage and current to real circuits. Although rectangular notation is convenient for addition and subtraction, and was indeed the final step in our sample problem here, it is not very applicable to practical measurements.

Rectangular figures must be converted to polar figures (specifically polar magnitude) before they can be related to actual circuit measurements. We can use SPICE to verify the accuracy of our results. In this test circuit, the 10 kΩ resistor value is quite arbitrary. It’s there so that SPICE does not declare an open-circuit error and abort analysis. Also, the choice of frequencies for the simulation (60 Hz) is quite arbitrary, because resistors respond uniformly for all frequencies of AC voltage and current. There are other components (notably capacitors and inductors) which do not respond uniformly to different frequencies, but that is another subject! Sure enough, we get a total voltage of 36.81 volts ∠ -20.5o (with reference to the 15 volt source, whose phase angle was arbitrarily stated at zero degrees so as to be the “reference” waveform). At first glance, this is counter-intuitive. How is it possible to obtain a total voltage of just over 36 volts with 15 volt, 12 volt, and 22 volt supplies connected in series?

With DC, this would be impossible, as voltage figures will either directly add or subtract, depending on polarity. But with AC, our “polarity” (phase shift) can vary anywhere in between full-aiding and full-opposing, and this allows for such paradoxical summing. What if we took the same circuit and reversed one of the supply’s connections? Its contribution to the total voltage would then be the opposite of what it was before: (Figure below) Polarity of E2 (12V) is reversed. Note how the 12 volt supply’s phase angle is still referred to as 35o, even though the leads have been reversed. Remember that the phase angle of any voltage drop is stated in reference to its noted polarity. Even though the angle is still written as 35o, the vector will be drawn 180o opposite of what it was before: (Figure below) Direction of E2 is reversed. The resultant (sum) vector should begin at the upper-left point (origin of the 22 volt vector) and terminate at the right arrow tip of the 15 volt vector: (Figure below)