Math 170 Proofs

Suppose X E R If x^3-X> 0 then x>-1

*Proof by contrapositive* x≤-1 and x^3-x≤0 Proof Let X E R Suppose for proof by contrapositive that x≤-1 This means x + 1 ≤ 0 It follows that x≤-1 and x-1≤-2 So multiplying x<0 with x-1<0 we get x(x-1) > 0 Multiplying the last inequality by the non-positive term x+1 to get x(x-1)(x+1) = x^3-x≤0 Completes proof by contrapositive

Prove that 1^2 + 2^2 + 3^2 + .... n^2 = [n(n+1)(2n+1)]/6 for every positive integer n

*Proof by mathematical induction* Proof base step for n=1 1^2= 1x2x3/6= 1 which is true so it work Assuming the statement is true for n=k *plug k into the equation* Then the statement MBT for K+1 *plug k+1 into the equation* to find that the right side equals L = [(k+1)(k+2)(2k+3)]/6 (R) *now you want to plug k+1 into the left side of the equation* --> you want the L = R k^2 + (k+1)^2 *make like terms over 6 *first get the (k+1) to the right *factor a k out then it gets rid of the square *(k+1)(2k^2 + 7k +6) / 6 = (k+1) (k+2) (2k+3) // 6 Left hand and right side equal This proves the inductive step Therefore by principle of mathematical induction, the given statement is true for every positive integer n

ch. 5 #8

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If X E R and 0 < X < 4 then (4|(x(4-x)) is ≥ 1

Proof Suppose X E R and 0<X<4 Since X-2 E R we have (X-2)^2 ≥ 0 So 0 ≥-x^2 + 4x + 4 4 ≥ X(4-X) 0<X and X<4 So X and 4-X are positive Thus you can divide the inequality by X(4-x) to get (4|(X(4-X)) ≥ 1

Suppose a is an integer. If 5|2a then 5|a

*Direct Proof* Proof Suppose 5|2a 2a=5x for some integer x 2a is even by definition so 5x must be even Thus x must be even x=2n 2a=5(2n) 2a=10n a=5n Thus 5|a by definition of divisibility

Suppose an and b are integers If a|b then a|3b^3-b^2+5b ????

*Direct Proof* Proof Suppose a|b b=ax for some integer x Thus 3(ax)^2 - (ax)^2 +5(ax) =3a^2x^2 -a^2x^2 +5ax =a (3ax^2 -ax^2 +5x) z=(3ax^2 -ax^2 +5x) a|az Therefore a|3b^3-b^2+5b

If x is an odd integer, then x^3 is odd

*Direct Proof* Proof Suppose x is odd, thus x=2a+1 for some a E Z x^3= (2a+1)(2a+1)(2a+1) x^3=4a^2+4a+1 (2a+1) x^3=8a^3+12a^2+6a+1 x^3= 2(4a^3 + 6a^2 + 3a) + 1 n= (4a^3 + 6a^2 + 3a) x^3=2n+1 Therefore x^3 is odd by definition of an odd integer

Suppose a,b,c E Z If a|b and a|c then a|(b+c)

*Direct Proof* Proof By definition of divisibility b=ae and c=af b+c=ae+af b+c = a(e+f) therefore by the definition of divisibility a|(b+c)

If n E z then 5n^2+3n+7 is odd

*Direct Proof* Proof suppose n E z Thus n is either even or n is odd *Even* n=2a for some integer a =5(2a)^2 + 3(2a) + 7 =10a^2 + 6a + 7 =2(5a^2 + 3a + 3) + 1 =2x + 1 Therefore 5n^2+3n+7 is odd *Odd* n=2a+1 for some integer a =5(2a+1)^2 + 3(2a+1) + 7 =5(4a^2+4a+1) + 6a+ 3 + 7 =20a^2 + 20a + 5 + 6a + 3 + 7 =20a^2 + 26a + 15 =2(10a^2 + 13a + 7) + 1 =2x + 1 Therefore 5n^2+3n+7 is odd

Suppose X,Y E Z If x and y are odd, then xy is odd

*Direct Proof* Proof x=2a+1 y=2a+1 xy= (2a+1)(2a+1) xy=4a^2 + 4a + 1 xy= 2(2a^2 + 2a) + 1 n = 2a^2 + 2a xy= 2n +1 Therefore xy is odd by definition of divisibility

If n ∈ N, then 1/2! + 2/3! + 3/4! + n/(n+1)! = 1 - 1/(n+1)!

*Mathematical induction* Proof Base Step: When n=1 the left hand side is 1/2!=1/2 When n=1 the right hand side is 1 - 1/(1+1)! = 1/2 Which proves the equality is true when n=1 Inductive Step: Let n≥1 We are assuming that the statement is true Our goal is to show 1/2! + 2/3! + 3/4! + n+1/((n+1)+1)! = 1 - 1/(n+1)! Add what we got from the first right side to the n+1 and then simplify so that you end up with the outcome as =1-(1/(n+1)+1)!) (see comp screenshot)

Suppose a, b, c E Z If a^2+b^2=c^2, then a or b is even

*Proof by contradiction* Proof Suppose for proof by contraction that there are integers a, b, and c for which a^2+b^2≠c^2 with BOTH a and b as odd integers Thus a=2n+1 with integer n E Z b=2m+1 with integer m E Z Thus c^2=a^2+b^2 = (2n+1)^2+(2m+1)^2 =4(n^2+n+m^2+m)+2 = C^2 C^2 - 4(n^2+n+m^2+m)+2 = 0 Contradiction because uses "If a,b E Z then a^2-4b-2 ≠ 0" where b=(n^2+n+m^2+m)

If a and b are positive real numbers then a+b≥2√ab

*Proof by contradiction* Proof Suppose for proof by contradiction that a and b are positive real numbers for which a+b<2√ab Since a+b > 0 squaring both sides (a+b)^2 < 4ab Subtract 4ab from both sides (a+b)^2 - 4ab < 0 a^2-2ab+b^2 < 0 (a-b)^2 < 0 Since (a-b) is a real number and the square of any real number is a non negative number there is a contradiction

If a,b E Z then a^2-4b-2 ≠ 0√

*Proof by contradiction* Proof Suppose for proof by contradiction that there are integers a and b for which a^2-4b-2 = 0 a^2= 4b+2 a^2=2(2b+1) Since 2b+1 E Z a^2 is even thus a is even a=2n for some integer n a^2=(2n)^2= 2(2b+1) (2n)^2= 2(2b+1) 4n^2 = 2(2b+1) --> (divide both sides by 2) 2n^2 = 2b+1 Since n^2 E Z the last equality shows 2b+1 is both even and odd Contradiction, proved

If A and B are sets, then A∩(B-A) = ∅

*Proof by contradiction* Proof Suppose for proof by contradiction that there are sets A and B such that A∩(B-A) ≠ ∅ Since A∩(B-A) is not an empty set is has an element X (by definition of an empty set) Thus X E A∩(B-A) Which means X E A and X E (B-A) Which means X E B and X is not included in A We have shown X E A and X is not included in A a contradiction

Prove that √6 is irrational

*Proof by contradiction* Proof Suppose √6 is rational This means there exists a,b E z where √6= a/b --> 6/(a^2/b^2) --> 6b^2=a^2 Which means 6|a^2 Which means 6|a a is divisible by 6, 6 is a factor of a a=6k, K E Z a^2 = 6b^2 -->6b^2= (6k)^2 --> 6b^2 = 36k^2 --> b^2 = 6k^2 b is divisible by 6 -->a and b are both divisible by 6 (they can't share factors) Therefore, √6 is irrational

Suppose n E z, If n^2 is odd, then n is odd

*Proof by contrapositive* Let n E z For sake of proof by contrapositive assume n is not odd Thus n is even n=2a for some integer a so n^2= (2a)^2 n^2= 2(2a^2) let x= 2a^2 n^2=2x Therefore n^2 is even Proof is complete

Suppose a, b, c E Z If a does not divide bc then a does not divide b

*Proof by contrapositive* a divides b, and a divides bc Proof Suppose a, b, c E z Suppose for proof by contrapositive a|b b= ax for some integer x So bc= axc Since xc E Z, this shows a|bc