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### assignment8_sol

Course: LINEAR ALG MATH135, Spring 2008
School: Waterloo
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Word Count: 2728

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135 MATH Assignment 8 Solutions Hand-In Problems Exercise 7-1: The following is known to be a simple substitution cipher. Break the code. PCTPG BTHHP TUDAA ANJHT VTHHT DLXCV GDURG CIIDW HXBEA NEIDV XHIGD TBTIW GPEWN DEHLT DS LPHYJ GTSXH AXJHR VJXHT PTHPG SQNIW Fall 2005 Solution: The given ciphertext has 102 letters. Analyzing it we see that the letter T appears 10 times, and this is one the most repeated...

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135 MATH Assignment 8 Solutions Hand-In Problems Exercise 7-1: The following is known to be a simple substitution cipher. Break the code. PCTPG BTHHP TUDAA ANJHT VTHHT DLXCV GDURG CIIDW HXBEA NEIDV XHIGD TBTIW GPEWN DEHLT DS LPHYJ GTSXH AXJHR VJXHT PTHPG SQNIW Fall 2005 Solution: The given ciphertext has 102 letters. Analyzing it we see that the letter T appears 10 times, and this is one the most repeated letters. Because the letter E is so common in the english language, an educated guess is to try the substitution cipher f that sends E to T. This corresponds to the function f : A A defined by f (x) x + 15 (mod 26), where A 1, B 2, . . .. The inverse of this function would be f -1 (x) x - 15 (mod 26). Applying f -1 to the message we get PCTPG ANEAR BTHHP MESSA TUDAA EFOLL ANJHT LYUSE VTHHT GESSE DLXCV OWING GDURG ROFCR CIIDW NTTOH HXBEA SIMPL NEIDV YPTOG XHIGD ISTRO TBTIW EMETH GPEWN RAPHY DEHLT OPSWE DS OD LPHYJ WASJU GTSXH REDIS AXJHR LIUSC VJXHT GUISE PTHPGCiphertext AESARPlaintext SQNIWCiphertext DBYTHPlaintext Ciphertext Plaintext With the correct spacing it would be " an early user of cryptography was Julius Caesar messages sent to his troops were disguised by the following simple method". Exercise 7-11: Given n = pq, p > q, and (n) = (p - 1)(q - 1) prove that p + q = n - (n) + 1 Solution: For the first part, n - (n) + 1 = pq - (p - 1)(q - 1) + 1 = pq - (pq - p - q + 1) + 1 = p + q. and p-q = (p + q)2 - 4n. For the second part, (p + q)2 - 4n = p2 + 2pq + q 2 - 4pq = p2 - 2pq + q 2 = (p - q)2 . Since p - q > 0, we have (p + q)2 - 4n = p - q. Exercise 7-14: Each integer n is the product of two primes p and q, and the Euler phi function (n) = (p - 1)(q - 1). Determine the prime factors p and q. n = 71531, (n) = 70992 Solution: From Exercise 7-11, p+q p-q = n - (n) + 1 = 71531 - 70992 + 1 = 540 = (p + q)2 - 4n = 5402 - 4(71531) = 74 Adding the two equations, 2p = 614 so p = 307. Since p - q = 74 and p = 307, then q = 233. Check: (233)(307) = 71531 and (232)(306) = 70992. Exercise 7-28: Encrypt each message M , using the RSA public key (e, n) and the square and multiply algorithm. M = 2041, (e, n) = (13, 3599) Solution: To encrypt the message M we have to calculate [M 13 ] in Z3599 . Using the square and multiply algorithm, 13 = (1101)2 = 8 + 4 + 1 and 20412 20414 20418 204113 1638 (mod 3599) 16382 1789 (mod 3599) 17892 1010 (mod 3599) = 20418 20414 2041 1010 1789 2041 192 2041 (mod 3599) 3180 (mod 3599) (mod 3599) So the enciphered message of M is C = 3180. Exercise 7-36: Use the Chinese Remainder Theorem to decrypt each received ciphertext C, using the RSA private key (d, n) where n = pq. C = 1120, d = 5051, p = 79, q = 131 Solution: We calculate C d modulo p and modulo q and then combine using the Chinese Remainder Theorem to get C d modulo n. Look first at C 5051 modulo 79. We can simplify the calculation by recognizing that since C 0 (mod 79), then C 78 1 (mod 79). This means that C 78q 1 (mod 79) for any integer q. Since 5051 = 64(78) + 59, then C 5051 (C 78 )64 C 59 C 59 (mod 79). This drastically reduces the power which we have to work with. Now we use the square and multiply algorithm to calculate 112059 modulo 79, simplifying this initially by recognizing that 1120 14 (mod 79). Since 59 = (111001)2 = 32 + 16 + 8 + 2 + 1, we have 142 144 148 1416 1432 So 1459 46 21 10 38 14 12 (mod 79) Similarly, d = 5051 = 38(131) + 111 and C = 1120 72 (mod 131) and so 111 = (1101111)2 = 64 + 32 + 8 + 4 + 2 + 1. We have 722 724 728 7216 7232 7264 75 (mod 752 123 1232 64 642 35 352 46 462 20 131) (mod 131) (mod 131) (mod 131) (mod 131) (mod 131) 38 382 222 102 212 (mod 22 10 21 46 79) (mod (mod (mod (mod 79) 79) 79) 79) So 72111 20 46 64 123 75 72 127 Thus, 1120 5051 (mod 131) 72 111 127 (mod 131). R 12 (mod 79) using the Chinese Remainder Theorem. R 127 (mod 131) When we apply the Chinese Remainder Theorem in the usual way, we obtain R 7201 (mod 10349), so R = M = 7201. Finally, we must solve the pair of congruences Exercise 7-40: Anne has a public key (eA , nA ) = (7, 8453), and a private key (dA , nA ) = (7087, 8453). Bill has public key (eB , nB ) = (1837, 9379), and private key (dB , nB ) = (5, 9379). Anne sends to Bill a signed message encrypted under Bill's public key. The ciphertext comes in two enciphered blocks. 5752 Find the message sent by Anne. Solution: If a, a-1 and b, b-1 are Anne's and Bill's public encryption and private decryption functions, respectively, then the signed messages M1 and M2 encrypted under Bill's public key are b(a-1 (M1 )) = 5752 and b(a-1 (M2 )) = 7155. To decipher them we must compute M1 = a(b-1 (5752)) and M2 = a(b-1 (7155)). Now b-1 (5752) = 5752dB in Z9379 . Using the square and multiply algorithm, 5 = 22 + 20 . 5752 5752 (mod 9379) 57522 57522 5771 (mod 9379) 57524 57712 8991 (mod 9379) 57525 57524 5752 (mod 9379) 8991 5752 (mod 9379) 426 (mod 9379) 7155. Because 426 < 8453, we continue. (If this was not true then we would have problems.) We now have to compute a(426) = 4267 in Z8453 . Using the square and multiply algorithm, 7 = 22 + 21 + 20 . 426 426 (mod 8453) 4262 4262 3963 (mod 8453) 4264 39632 8148 (mod 8453) 4267 4264 4262 426 (mod 8453) 8148 3963 426 (mod 8453) 1905 (mod 8453) Now we do the same thing for 7155. 7155 7155 (mod 9379) 71552 71552 3443 (mod 9379) 71554 34432 8572 (mod 9379) 71555 71554 7155 (mod 9379) 8572 7155 (mod 9379) 3379 (mod 9379) 3379 < 8453, so we continue: 3379 3379 (mod 8453) 33792 33792 6091 (mod 8453) 33794 60912 64 (mod 8453) 33797 33795 33792 3379 (mod 8453) 64 6091 3379 (mod 8453) 1212 (mod 8453) Therefore the original message was 1905 1212, so using the equivalence A 01 etc. we get 19051212 SELL. Problem 7-48: (a) Let n = pq, where p and q are primes. Prove that if p - q is known then n can be factored.(Exercise 7-11 will help.) (b) Describe a way to break an RSA system with modulus n = pq if p - q is not too large. (This problem illustrates an important point. When constructing an RSA system one must pick the primes so that their difference is large.) Solution: (a) Suppose, p - q is known. We also know n. By Exercise 7-11, p-q (p - q)2 (p + q)2 p+q = (p + q)2 - 4n = (p + q)2 - 4n = (p - q)2 + 4n = (p - q)2 + 4n Since we know p + q and p - q, then from these, we can determine p = (1/2) [(p - q) + (p + q)] and q = n - p. (b) If (p - q) is not too large, we can, by part (a), find it by testing small values of i to check whether i2 + 4n is a perfect square (ie. look at i = 1, 2, 3, 4, . . .). (Here, we are letting i represent our potential values for p - q and trying to see if we can get an integer value for p + q. Knowing p - q and p + q allows to solve for p and q.) Then let p - q = i0 , where i0 P is such that i2 + 4n is the square of an integer, and let p + q = 0 Determine p and q as described in (a) and compute (p - 1)(q - 1). Then, find the private key d by solving ed 1 (mod (p - 1)(q - 1)). i2 + 4n. 0 This only works if (p - q) is relatively small, because then only relatively few i P need to be tested. If (p - q) is large, the testing will take too long to be useful. Exercise 8-18: Express each of the following complex numbers in the standard form x + iy, where x, y R. 1 i65 Solution: Because 65 1 (mod 4) then by Proposition 8.23, i65 = i, so 1 1 -i = = = 0 + i(-1) = -i. i65 i 1 Thus, 1 i65 = 0 + i(-1). 1 = i-65 and noticed that -65 3 (mod 4).) i65 (Alternatively, we could have written Exercise 8-22: Express each of the following complex numbers in the standard form x + iy, where x, y R. ( 2 - i)2 ( 2 + i)(1 - 2i) Solution: ( 2 - i)2 ( 2 + i)(1 - 2i) = = = = (1 - 2 2i) (1 - 2 2i) 2 2 + i = (2 2 - i) (2 2 - i) 2 2 + i (1 - 2 2i)(2 2 + i) 9 4 2 - 7i 9 4 2 -7 +i . 9 9 Exercise 8-32: What the is complex conjugate and modulus of each of the following numbers? - 3-i 7 Solution: If z = - 3 - i 7, then z = - 3 + i 7 and |z| = (- 3)2 + (- 7)2 = 10. Exercise 8-36: Plot the following points in the complex plane. z = 1 + i, z 2 , z 3 , z 4 and 1/z Solution: If z = 1 + i then z2 z3 z4 1 z = (1 + i)(1 + i) = 1 + 2i + i2 = 2i = z 2 z = (2i)(1 + i) = -2 + 2i = z 2 z 2 = (2i)(2i) = -4 1 1-i 1-i 1 1 = = = = -i 1+i (1 + i)(1 - i) 2 2 2 If we plot these points in the complex plane we have, 6 zr3 z 2 r d rz di d r 2 1/z 4 zr - ? Exercise 8-40: Find polar coordinates for each of the following points that are given in Cartesian form. (- 3, 1) Solution: First we find the modulus: |(- 3, 1)| = (- 3)2 + (1)2 = 2. Next we find the argument. If is the argument of (- 3, 1) then cos = -2 3 . Since the point lies in the second quadrant, then = . 5 6 So (- 3, 1) in polar coordinates is (2, 5/6). Check: x = 2 cos (5/6) = - 3 and y = 2 sin (5/6) = 1. Exercise 8-44: Each of the following pairs of numbers gives the polar coordinates of a point in the plane. Express each point in Cartesian coordinates. (7, 7/6) Solution: x = 7 cos(7/6) = -7 3/2 and y = 7 sin(7/6) = -7/2. So the point in polar coordinates (7, 7/6) corresponds to point in the Cartesian coordinates (-7 3/2, -7/2). Exercise 8-54: Express each of the following numbers rcis = r(cos + i sin ) in the standard form x + iy where x, y R. 3 cis(4/3) Solution: x = 3 cos(4/3) = - 3/2 and y = 3 sin(4/3) = -3/2. Hence, 3 cis 4 = - 23 - i 3 . 3 2 Exercise 8-58: Find the modulus and argument of each of the following complex numbers. 2 + (3 - 5)i Solution: The modulus is | 2 + (3 - 5)i| = ( 2)2 + (3 - 5)2 = 16 - 6 5 1.61. 5 > 0) and makes an angle with the positive x- The number is in the first quadrant (since axis that is given by cos = 2 . 16-6 5 2 > 0 and 3 - Thus the argument is 0.495 in radians. Recommended Problems Exercise 7-3: For each of the following values of p, q, and e, find the public key (e, n), and the associated private key (d, n) of an RSA scheme. p = 17, q = 19, e = 25. Solution: We calculate n = pq = 323 and (p - 1)(q - 1) = 288 = 25 32 . Note that gcd(e, (n)) = gcd(25, 288) = 1, so we have to solve the congruence 25d 1 (mod 288). The congruence is equivalent to the Diophantine Equation 25y + 288x = 1 for x, y Z. By inspection (or otherwise), 25(-23) + 288(2) = 1. So d -23 265 (mod 288) is the unique solution. The public encryption key is (e, n) = (25, 323) and the private key is (d, n) = (265, 323). Check: ed = (265)25 = 6625 = 23(288) + 1. Exercise 7-7: For each of the following public keys (e, n), determine the associated private key (d, n) of an RSA scheme. (e, n) = (5, 7663) Solution: We need the prime factorization of 7663 to find (p - 1)(q - 1), and hence find d. Because n is small it can be factored trying prime factors less than 7663 87.5. (Of course, normally we couldn't factor n in this way.) After some trial and error, we find that 7663 = 79(97) and 79 and 97 are indeed prime. Then (p - 1)(q - 1) = 7488 = 26 32 13 and gcd(5, 7488) = 1. Now we have to solve the congruence 5x 1 (mod 7488) which is equivalent to the Linear Diophantine Equation 5x + 7488y = 1 for x, y Z. Using the Extended Euclidean Algorithm 7488y + 5x = r qi 1 0 7488 0 1 5 1 -1497 3 -1 1498 2 2 -2995 1 1497 1 1 So 5(-2995) + 7488(2) = 1 and d -2995 4493 (mod 7488) is the unique solution to the congruence. Therefore, the public encryption key is (e, n) = (5, 7663) and the private key is (d, n) = (4493, 7663). Check: ed = (5)4493 = 224468 = 3(7488) + 1. Exercise 7-16: (a) Prove that the encryption function f : Zn Zn , for the RSA system defined by f [x] = [xe ], is a bijection. (b) Find the permutation of Z15 defined by the bijection f : Z15 Z15 with f [x] = [x7 ]. Solution: (a) The functions f : Zn Zn , and g : Zn Zn defined by f ([x]) = [xe ] and g([y]) = [y d ] are inverse functions. (We know this from the set-up of RSA.) This implies that f is invertible and by the Inversion Theorem f is a bijection (since f is invertible if and only if it is a bijection). (b) f : Z15 Z15 defines the permutation, 1 2 f (1) f (2) = 0 0 1 1 2 8 3 12 4 4 14 15 f (14) f (15) 5 5 6 6 7 13 8 2 9 10 9 10 11 11 12 3 13 7 14 14 (This is really a table which lists the values of f ([x]) for each class in Z15 .) Exercise 7-17: Encrypt each message M , using the RSA public key (e, n). M = 47, (e, n) = (5, 119) Solution: To encrypt the message M we have to calculate [M 5 ] in Z119 . We use the square and multiply algorithm. Because 5 = 22 + 20 and 472 67 (mod 119) and 474 86 (mod 119), then 475 86 47 115 (mod 119). So the encrypted message is 115. Exercise 7-21: Decrypt each received ciphertext C, using the RSA public key (d, n). C = 32, (d, n) = (77, 119) Solution: To decrypt each ciphertext we must calculate [C 77 ] in Z119 . We use the square and multiply algorithm. Now, 77 = 64 + 8 + 4 + 1 = 26 + 23 + 22 + 20 so C C2 C4 C8 C 16 C 32 C 64 C 77 32 322 722 672 862 182 862 (mod 72 67 86 18 86 18 119) (mod (mod (mod (mod (mod (mod 119) 119) 119) 119) 119) 119) C 64 C 8 C 4 C (mod 119) 18 86 72 32 (mod 119) 2 (mod 119) Exercise 7-31: For each exponent e, determine the number of modular multiplications to encrypt an RSA message using the square and multiply algorithm. e = 92487. Solution: After some work, 92487 = 216 + 214 + 213 + 211 + 28 + 26 + 22 + 21 + 20 = (10110100101000111)2 . k Thus we need 16 squarings to calculate the remainder of M 2 modulo n for k = 1 to k = 16, and 8 multiplications for a total of 24 modular multiplications. Exercise 7-39: Let (e, n) be the public encryption key for an RSA system. Suppose that it takes 10-4 seconds to do one modular multiplication, and that e has 200 digits with 100 ones in its binary representation. Assuming that modular multiplication is the only time-consuming operation, determine the time required to encrypt a message. Solution: Let e = (r200 r199 . . . r2 r1 r0 )2 where each ri = 0 or 1. Because the binary representation of e has 200 digits we know that r200 = 1 so e = 2200 + . k Therefore 200 squarings are needed in order to calculate the remainder of M 2 for k all the way up to 200. And because there are 100 ones in the binary representation of e, 100 of these powers must be multiplied. This involves 99 extra multiplications for a total of 299 modular multiplications. Hence the time required is t = ( # of multiplications) (time per multiplication) = 299 10-4 s = 0.0299 seconds Problem 7-45: We would like to distribute some information to s people so that if any two of the s people combine their information they can deduce the secret positive integer k, but no person alone can do so. Select a prime number p larger than s and k. Then select a polynomial f (x) = ax + k Zp [x] with a = 0. Compute pairs (i, f (i)), 1 i s, and distribute these to the s people. Prove that this scheme has the desired properties. Solution: Clearly p = 2 because p > s 2. If two people have (i, f (i)) and (j, f (j)), 1 i, j s p and i = j, then they can calculate f (i) - f (j) = (ai + k) - (aj + k) = a(i - j). Because i - j 0 (mod p) then it has an inverse in Zp and because both people know i and j they can calculate it by solving the congruence (i - j)x 1 (mod p). Multiplying f (i) - f (j) by it gives (i - j)-1 (f (i) - f (j)) = a. Knowing a, either of the two can calculate k = f (i) - ai = f (j) - aj. Now we have to prove that one person alone cannot find out k, say person i. Because a = 0 and p = 2, there exists d Zp such that d = 0 and a + d = 0 and k - di k (mod p). Consider g(x) = (a + d)x + (k - di). It is clear that f (i) = g(i) but f and g have different constant terms. Therefore one person alone cannot determine a unique k from (i, f (i)). Exercise 8-23: What are the real and imaginary parts of the following complex numbers? 4 - 7i Solution: If z = 4 - 7i then Re(z) = 4 and Im(z) = -7. Exercise 8-25: What are the real and imaginary parts of the following complex numbers? -3 Solution: If z = -3 = -3 + 0i, then Re(z) = -3 and Im(z) = 0. Exercise 8-51: Express each of the following complex numbers in their polar form. i 1 + . 3 3 Solution: We have |(1/ 3) + (i/ 3)| = 1 3 2 + 1 3 2 = 2 3 = 6 3 . The point lies in the first quadrant, since x > 0 and y > 0. Since x = y, the argument is = /4. 1 i Hence, 3 + 3 = 36 cos + i sin . 4 4 Exercise 8-60: Show that (1 + i2n )(1 + in ) is either 0 or 4 for n P. Describe the values of n that yield the various answers. Solution: We have (1 + (i2 )n )(1 + in ) = (1 + (-1)n )(1 + in ). By Proposition 8.23, we only need to check the congruence of n modulo 4 in order to determine the value of in . n 0 (mod 4) n 1 (mod 4) n 2 (mod 4) n 3 (mod 4) in in in in =1 =i = -1 = -i and and and and (-1)n (-1)n (-1)n (-1)n =1 = -1 =1 = -1 So So So So (1 + (i2 )n )(1 + in ) = 2(2) = 4 (1 + (i2 )n )(1 + in ) = 0(1 + i) = 0 (1 + (i2 )n )(1 + in ) = 2(0) = 0 (1 + (i2 )n )(1 + in ) = 0(1 - i) = 0 Therefore (1 + i2n )(1 + in ) is 0 for n 1, 2, 3 (mod 4) and is 4 for n 0 (mod 4). Exercise 8-61: Solve the equation z = z 2 for z C. Solution: Let z = x + iy for x, y R. Then z = x - iy. If z = z 2 then x - iy = (x2 - y 2 ) + i(2xy). This happens if and only if x = x2 - y 2 and -y = 2xy. If y = 0 the second equation yields 2x = -1 so x = -1/2. Substituting in the second equation gives y 2 = 3/4 so y = 3/2. If y = 0 then x2 = x which implies x = 0 or 1. So we have four solutions z = -1/2 i( 3/2), z = 0 + i(0) and z = 1 + i(0).
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1 Timothy. Theme: order in the church Teachings. A. Warnings of false teachers (Jewish) B Pray for our authority. C. Woman (Modest Dress) 1. Jewelry, braided hair, expensive clothes. D. Woman in worship 1. Learn in quietness 2. Not to take authority
Oklahoma Christian - EDU - literacy 1
MODEL LESSON PLAN FORMAT Aligned with Oklahoma's Minimum Criteria for Effective Teaching and Teacher Work Sample SUBJECT MATTER _ _ Instructional Time/Period of the Day _DATE(s) _ Essential Elements of Planning Elements Notes about theEstablish Con
Oklahoma Christian - EDU - LIA
Kimberly Fletcher Early Childhood(LIA) Factor TwoOklahoma Pass Standard 1; Print Awareness - The student will understand the characteristics of written language. Pass skill; 1. Demonstrate correct book orientation by holding book correctly (right
Oklahoma Christian - EDU - movement i
Kimberly FletcherStudio DramaAge Level Curriculum Area Objectives Adaptations For Special NeedsAdaptation for different age levels MaterialsDirectionsSource Extensions4th grade Science The kids will be able to give facts about the process
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementFlower walkAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials Directions5th Social Studies 1. Students will be able to recognize the state flower
Oklahoma Christian - EDU - movement i
So, who was Saint Valentine and how did he become associated with this ancient rite? Today, the Catholic Church recognizes at least three different saints named Valentine or Valentinus, all of whom were martyred.
Oklahoma Christian - EDU - movement i
Category: Visual Arts Age Level Curriculum Area ObjectivesRhyming WordsKimberly FletcherAdaptations For Special Needs Adaptation for different age levelsMaterialsDirectionsSource ExtensionsKindergarten Language arts The students will be
Oklahoma Christian - EDU - movement i
Category: Visual Arts Age Level Curriculum Area Objectives Adaptations For Special Needs7's FamilyKimberly FletcherAdaptation for different age levelsMaterialsDirectionsSource Extensions3rd Grade Math Students will be able to write thei
Oklahoma Christian - EDU - movement i
Kimberly FletcherDrama StudioAge Level Curriculum Area Objectives Adaptations For Special NeedsAdaptation for different age levelsMaterials DirectionsSource Extensions4th grade Science The students will be able to identify characteristics
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementJump the creekAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials2nd Math 1. The student will learn to measure distance. 2. The students will learn
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementInterpretive danceAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials DirectionsSource Extensions4th History 1. The students will have more knowl
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio LiteraryScience and litAge Level Curriculum Area ObjectivesAdaptations For Special Needs4th Language arts and Science 1. Students will be able to give key about the scientific method 2. Students will gain creative wri
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementDigestive dance party.Age Level Curriculum Area Objectives4th Science 1. Students will better understand the digestive system. 2. Students will be able to move around and release energy. 3. Students will have a
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio Visual ArtBird WatchAge Level Curriculum Area ObjectivesAdaptations For Special Needs1st Science 1. Students will be able to recognize state birds. 2. Students will learn listening skills. 3. Students will learn to gi
Oklahoma Christian - EDU - movement i
Category: Visual Arts Age Level Curriculum Area Objectives Adaptations For Special Needs Adaptation for different age levelsCloudsKimberly FletcherMaterials DirectionsSource Extensions4th Grade Science The students will be able to identify
Oklahoma Christian - EDU - movement i
Category: Visual Art Age Level Curriculum Area Objectives Adaptations For Special Needs Adaptation for different age levelsPilgrimsKimberly FletcherMaterialsDirectionsSource Extensions3rd Grade History The students will be able to identif
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MusicDigeridooAge Level 3rd Curriculu culture m Area Objectives 1. Students will be able to understand the musical culture of Australia 2. Students will be able to be creative with their hands 3. Students will learn to fo
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementPlanet SignAge Level Curriculum Area Objectives2nd Science1. Students will have knowledge of the planets 2. The students will learn to sign their planets. 3. The students will learn listening skills. 4. Studen
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementMath jumping jacksAge Level Curriculum Area Objectives1st Math 1. Students will learn their addition tables 2. Students will learn listening skills. 3. Students will be able to move around and get the wiggles ou
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementMath raceAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterialsDirections4th Math 1. Students will be able to Correctly answer the division problems
Oklahoma Christian - EDU - intro to e
Kimberly Thompson Philosophy ReflectionI believe that my philosophy paper shows that I am prepared for the teaching profession. I have shown that I have the desire and love for teaching. I have also shown that I have had experience in the field. I
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MusicAlphabet scrambleAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials Directions2nd Language art 1. Students will be able to put things in alphabetical
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MovementRibbon DanceAge Level Curriculum Area ObjectivesK5 Space around Us and science1. Students will understand what the space around us is. 2. Students will be able to understand a small idea of gravity. 3. Student
Oklahoma Christian - EDU - movement i
Kimberly Fletcher Studio MusicSpelling movement storyAge Level Curriculum Area ObjectivesAdaptations For Special NeedsAdaptation for different age levelsMaterials DirectionsSource Extensions3rd Spelling 1. The students will be able to c
Oklahoma Christian - EDU - movement i
Kimberly FletcherLiterary Studio5thAge Level Curriculum Area Objectives Adaptations For Special Needs Adaptation for different age levelsMaterials DirectionsSource Extensionsgrade history The students will be able to recite their states. I
Oklahoma Christian - BIBL - New testim
Word Identification Strategies. PG 134-171 1. Letter Sound Bingo- Read words spelled with the same letter sound patterns children are learning. B. Interactive bulletin boards- In a small group sort words according to shared letter sound patterns and
Oklahoma Christian - EDU - LIA
Kimberly Fletcher Factor 1 Literacy 1 practicum. Contextual Factor Number of Students;22 Female;11 Male;11 Ethnic and Cultural make-up; Sam is a white male.Socio economic status. and make up The school seemed to be average. The homes in the area se
Oklahoma Christian - EDU - LIA
Kimberly FletcherFactor 3 Post Assessment 2. Students will be able to recognize rhyming words from a list of given words.Draw a line to the correct rhyming word.1. 2. 3. 4. 5. 6. 7. 8. 9.Hall Stop Make Bite Man Sea Light Keep ToreDrop Can F
Oklahoma Christian - EDU - LIA
Kimberly FletcherFactor 3 Post Test Key1. Fall 2. Drop 3. Take 4. Fight 5. Can 6. Tea 7. Tight 8. Leap 9. Door
Oklahoma Christian - EDU - LIA
Kimberly Fletcher Factor 3 Type of Assessment 1. Pre Assessment The assessment specifically addresses the correct way to hold and read from a book. The test is reliable and valid because it shows that the student has mastered the challenge of holding
Oklahoma Christian - EDU - LIA
Kimberly Thompson Factor 1 Reflection Factor on has helped me in many ways. It has taught me how to evaluate my classroom. I learned that it is crucial to be familiar with my class makeup in order to be an effective teacher. I believe that this paper
Oklahoma Christian - EDU - movement i
Rhyming game activity.Like to rhyme? Well, now's the time! This game is for 2 or more players. You'll need pencils or pens and paper. You can play one on one or you can split into two even teams. You have one minute to think of as many pairs of wor
Marietta - COMM - 101
Informative Outline Name: Date: 9/20/2007 Title: African Conflict Diamonds Specific Purpose: To inform my audience about African conflict diamonds and the problems associated with these precious stones. Preview: While learning about African conflict