please answer the questions. it is (introduction to logic for computer science)if you do not answer all of them just give me what you know it is fine

1090-ass-6 Jeff Edmonds York University Assignment 4 MATH1090 Propositional Logic 1alpha 3,2 3 2alpha 2,3 3 3form vector 10 4prove general 20 5Interpolating 15 6song 15 7fft graphs 15 8Physics colour 4 9brain colour 15 total 100 I hope this will be • fun • educational • doable • not too scary. Linear Algebra In high school and first year linear algebra, you would have learned about • Euclidean Spaces • Points • Linear equations • A system of d-equations and d-unknowns leading to a unique solution 4 ´x1 + (-2)´x2 + 7 ´x3 = 3 (-8)´x1 + 5 ´x2 + (-3)´x3 = 12 9 ´x1 + 1 ´x2 + 6 ´x3 = -6 x1 x3 x2 x3 x1 x2 áx1,x2,x3ñ áx1,x2,x3ñ = á501/281, 141/281,447/281ñ Linear Algebra In high school and first year linear algebra, you would have learned about • Euclidean Spaces • Points • Linear equations • A system of d-equations and d-unknowns leading to a unique solution 4 ´x1 + (-2)´x2 + 7 ´x3 = 3 (-8)´x1 + 5 ´x2 + (-3)´x3 = 12 9 ´x1 + 1 ´x2 + 6 ´x3 = -6 x1 x3 x2 x3 x1 x2 áx1,x2,x3ñ áx1,x2,x3ñ = á501/281, 141/281,447/281ñ Linear Algebra 4 ´x1 + (-2)´x2 + 7 ´x3 = 3 (-8)´x1 + 5 ´x2 + (-3)´x3 = 12 9 ´x1 + 1 ´x2 + 6 ´x3 = -6 x1 x3 x2 x3 x1 x2 áx1,x2,x3ñ áx1,x2,x3ñ = á501/281, 141/281,447/281ñ • The solution áx1,x2,x3ñ could be any where in the 3-dimmional volume. • One equation narrows the possibilities to 2-dimmional plane. • Two equation narrows the possibilities to 1-dimmional line. • Three equation narrows the possibilities to 0-dimmional point. • Unless these equations are linearly dependent In which case they repeat information so we don’t have enough info. Linear Algebra 4 ´x1 + (-2)´x2 + 7 ´x3 = 3 (-8)´x1 + 5 ´x2 + (-3)´x3 = 12 9 ´x1 + 1 ´x2 + 6 ´x3 = -6 • Other useful notations are: aá1,1ñ´x1 + aá1,2ñ´x2 + aá1,3ñ´x3 = b1 aá2,1ñ´x1 + aá2,2ñ´x2 + aá2,3ñ´x3 = b2 aá3,1ñ´x1 + aá3,2ñ´x2 + aá3,3ñ´x3 = b3 Fori=1..d, ∑!"#..% aái,jñ´xj = bi [ ][ ] =[ ]4 (-2) 7 3(-8) 5 (-3) 129 1 6 -6x1x2x3 [ ][ ] =[ ]x1x2x3aá1,1ñ aá1,2ñ aá1,3ñ b1aá2,1ñ aá2,2ñ aá2,3ñ b2aá3,1ñ aá3,2ñ aá3,3ñ b3 Linear Algebra • Matrix Multiplication: • Each row is an equation • Dot produce of row and column. Sum row ´ column aá1,1ñ´x1 + aá1,2ñ´x2 + aá1,3ñ´x3 = b1 aá2,1ñ´x1 + aá2,2ñ´x2 + aá2,3ñ´x3 = b2 aá3,1ñ´x1 + aá3,2ñ´x2 + aá3,3ñ´x3 = b3 [ ][ ] =[ ]x1x2x3aá1,1ñ aá1,2ñ aá1,3ñ b1aá2,1ñ aá2,2ñ aá2,3ñ b2aá3,1ñ aá3,2ñ aá3,3ñ b3i j i i=2 Fori=1..d, ∑!"#..% aái,jñ´xj = bi j Linear Algebra • It is also useful to group everything having to do with the same i. aá1,1ñ´x1 + aá1,2ñ´x2 + aá1,3ñ´x3 = b1 aá2,1ñ´x1 + aá2,2ñ´x2 + aá2,3ñ´x3 = b2 aá3,1ñ´x1 + aá3,2ñ´x2 + aá3,3ñ´x3 = b3 [ ][ ] =[ ]x1x2x3aá1,1ñ aá1,2ñ aá1,3ñ b1aá2,1ñ aá2,2ñ aá2,3ñ b2aá3,1ñ aá3,2ñ aá3,3ñ b3 Fori=1..d, ∑!"#..% aái,jñ´xj = bi Each of these columns is also a “vector”. Hint: You will have to do this in one question. Linear Algebra [ ][ ] =[ ]x1x2x3aá1,1ñ aá1,2ñ aá1,3ñ b1aá2,1ñ aá2,2ñ aá2,3ñ b2aá3,1ñ aá3,2ñ aá3,3ñ b3 • Solving for x: • M x = b • x = M-1 b Dude, what’s all this algebra. We are here for logic!!! M x = b Let αalgebra be the following axiom: " real values (That are linearly independent) $ a tuples of real numbers áx1,x2,x3ñ, such that aá1,1ñ´x1 + aá1,2ñ´x2 + …+ aá1,dñ´xd = b1 aá2,1ñ´x1 + aá2,2ñ´x2 + …+ aá2,3ñ´xd = b2 … aá3,1ñ´x1 + aá3,2ñ´x2 + …+ aád,dñ´xd = bd Linear Algebra All you need to know about algebra is the validity of this axiom. aá1,1ñ aá1,2ñ … aá1,dñ b1 aá2,1ñ aá2,2ñ aá2,dñ b2 … aá3,1ñ aá3,2ñ aá3,dñ bd You will never need to prove independence. Just say “Clearly independent” Let αalgebra ≡ "[aái,jñ]&[bj], $áxjñ, [aái,jñ]´áxjñ = [bj] Linear Algebra All you need to know about algebra is the validity of this axiom. Matrix times a vector.Matrix Vectors Spanning Vector Spaces • In general, a vector v is an abstract object. But for now, let’s assume that it is • a tuple of real values v = á3,4,-8ñ • a location of a point in a room • or an arrow in some direction and of some length. • A vector space V is a set of such vectors. • An example of a vector space of dimension 3 is the set of all the points/vectors in your 3-dim room. • An example of a vector space of dimension 2 is the set of all the points/vectors on your 2-dim floor. • We add vectors as follows. v w v+w v = á3,4ñ w = á4,-1ñ v+w = á7,3ñ x1 x3 x2 x3 x1 x2 áx1,x2,x3ñ x1 x3 x2 x3 x1 x2 áx1,x2,x3ñ Spanning Vector Spaces v The standard basis for the Euclidian Vector space is = á1,0,0ñ = á0,1,0ñ = á0,0,1ñ. v = á3,4ñ Spanning Vector Spaces • The standard basis is the following tuple of d vectors: áu1,u2,…,udñ = á , , ñ = á á1,0,0ñ, á0,1,0ñ, á0,0,1ñ ñ. • A tuples of real numbers c = ác1,c2,…,cdñ = á3,4,-8ñ is just that. • Please do not confuse it with the vector v = á3,4,-8ñ = • These numbers ác1,c2,…,cdñ are often called coefficients. • A linear combination of the basis vectors is v = ∑&"#..% ci´ui = c1´u1 + c2´u2 + … + cd´ud = 3´á1,0,0ñ + 4´á0,1,0ñ + … + (-8)´á0,0,1ñ = á3,4,-8ñ v v = á3,4ñ c = á3,4ñ c2 = 4 c1 = 3x1 x3 x2 x3 x1 x2 áx1,x2,x3ñ Let αstandard(d´,d) be the following statement: " d´ dimensional vector spaces V, Let áu1,u2,…,udñ = [ , , … ] be V’s “standard basis”. " vectors vÎV, $ a unique tuple of real numbers ác1,c2,…,cdñ, such that v = c1´u1 + c2´u2 + … + cd´ud We “proved” this (by picture) when d´=d. Spanning Vector Spaces v v = á3,4ñ c = á3,4ñ c2 = 4 c1 = 3x1 x3 x2 x3 x1 x2 áx1,x2,x3ñ Note that vector v=v´ when they are equal in each coordinate. Spanning Vector Spaces Question 1: Prove or disprove α(3,2). Let αstandard(d´,d) be the following statement: " d´ dimensional vector spaces V, Let áu1,u2,…,udñ = [ , , … ] be V’s “standard basis”. " vectors vÎV, $ a unique tuple of real numbers ác1,c2,…,cdñ, such that v = c1´u1 + c2´u2 + … + cd´ud Spanning Vector Spaces Question 2: Prove or disprove α(2,3). Let αstandard(d´,d) be the following statement: " d´ dimensional vector spaces V, Let áu1,u2,…,udñ = [ , , … ] be V’s “standard basis”. " vectors vÎV, $ a unique tuple of real numbers ác1,c2,…,cdñ, such that v = c1´u1 + c2´u2 + … + cd´ud This becomes much more fun when we use a non-standard basis! • A non-standard basis is an arbitrary tuple of d vectors: áw1,w2,…,wdñ = á , , ñ = á á3,-1,2ñ, á2,2,8ñ, á-1,-1,3ñ ñ. • They must be “linearly independent” meaning that one isn’t the linear combination of the others. • Consider a tuple of real values áe1,e2,…,edñ = á3,4,-8ñ. • The linear combination of the basis vectors is v = ∑&"#..% ei´wi = e1´w1 + e2´w2 + … + edwd = 3´á3,-1,2ñ + 4´á2,2,8ñ + … + (-8)´á-1,-1,3ñ = á9,-3,6ñ + á8,8,32ñ + … + á8,8,-24ñ = á25,15,14ñ Spanning Vector Spaces You will never need to prove independence. Just say “Clearly independent” Spanning Vector Spaces Question 3: Now suppose: • V is your floor, • áw1,w2ñ = á , ñ. • v is • Give the (unique) tuple of real values áe1,e2ñ such that v = ∑&"#..% ei´wi = e1´w1 + e2´w2 Power point makes it easy to copy and move vectors without changing their length or direction. Let’s generalize this: Let αgeneral be the following statement: " d dimensional vector spaces V, " basis of linearly independent vectors áw1,w2,…,wdñ, " vectors vÎV, $ a unique tuple of real numbers áe1,e2,…,edñ, such that v = e1´w1 + e2´w2 + … + ed´wd Spanning Vector Spaces Matrix times a vector.Matrix Vectors Logic αalgebra ≡ "[aái,jñ]&[bj], $áxjñ, [aái,jñ]´áxjñ = [bj] αalgebra ≡ "[aái,jñ]&[bj], $áxjñ, [aái,jñ]´áxjñ = [bj] αstandard ≡ "V with standard basis áuiñ, "vÎV, $áciñ, v = áciñ‧áuiñ Dot product áciñ‧áuiñ = c1´u1 + c2´u2 + … + cd´ud Vector Space d vectors vector tuple of reals Logic αalgebra ≡ "[aái,jñ]&[bj], $áxjñ, [aái,jñ]´áxjñ = [bj] Logic αstandard ≡ "V with standard basis áuiñ, "vÎV, $áciñ, v = áciñ‧áuiñ αgeneral