proof

The square of a number is always nonnegative. This is true, and here is why: Suppose you have a positive number n. n^2 = n * n A positive times a positive is a positive Suppose you have a negative number -n (-n)^2 = -n * -n = n^2 A negative times a negative is a positive.

A U ∅ = A Let x ∈ S, where S is the universal set. First we show that if A ∪ Ø ⊂ A. Let x ∈ A ∪ Ø. Then x ∈ A or x ∈ Ø. by definition of the empty set, x cannot be an element in Ø. So by assumption, x ∈ A ∪ Ø, x must be in A. So A ∪ Ø ⊂ A. Next, we show that A ⊂ A ∪ Ø. This is true...

Let n be an integer. If n^2 is odd, then n is odd Proof by contraposition: Suppose that n is even. Then we can write n = 2k n^2 = (2k)^2 = 4k^2 = 2(2k) so it is even So an odd number can't be the square of an even number. So if an odd number is a square it must be the square of an odd number.

Given: 3(2x − 5) = 15 Prove: x = 5 Set x = 5: 3(2(5) - 5) 3(10 - 5) 3(5) 15 So if x = 5, then: 3(2x − 5) = 15

M is the midpoint of AB. Prove AB = 2AM M is the midpoint of AB (Given) AM = MB (Definition of Congruent Segments) AM + MB = AB (Segment Addition Postulate) AM + AM = AB (Substitution Property of Equality) 2AM = AB (Distributive property)

Let x be an integer. If x is odd, then x^2 is odd Proof: Let x be an odd number. This means that x = 2n + 1 where n is an integer. Squaring x, we get: x^2 = (2n + 1)^2 = (2n + 1)(2n + 1) x^2 = 4n^2 + 4n + 1 x^2 = 2(2n^2 + 2n) + 1 2(2n^2 + 2n) is an even number since 2 multiplied by any...

Given: WS bisects <RWP. Prove: <RWS=<PWS WS bisects <RWP (Given) <RWS = <PWS (Angle bisector Theorem)

Given: BC = EF AC = EG AB = 10 BC = 3 Prove FG = 10 AC = AB + BC (Segment Addition Postulate) AB = 10, BC = 3 (Given) AC = 10 + 3 (Substitution Property of Equality) AC = 13 (Simplify) AC = EG, BC = EF (Given) EG = 13, EF = 3 (Segment Equivalence) EG = EF + FG (Segment Addition Postulate)...

Given: <RST ~ <WSU Prove: m<RSW = m<TSU <RST < WSU. (Given) <RSW = <RST + <TSW (Angle Addition Postulate) <TSU = <TSW + <WSU (Angle Addition Postulate) Since <RST = <WSU (Given), we have: <RSW = <WSU + <TSW Therefore, <RSW = <TSU

n and m are congruent and supplementary. prove n and m are right angles Given: n and m are congruent n and m are supplementary If n and m are supplementary, that means we have the equation: m + n = 180 We're also given n and m are congruent, meaning they are equal. So we can substitute n = m...

Look at the Contrapositive: If n is even, then 3n + 2 is even... Suppose that the conclusion is false, i.e., that n is even. Then n = 2k for some integer k. Then we have: 3n + 2 = 3(2k) + 2 3n + 2 = 6k + 2 3n + 2 = 2(3k + 1). Thus 3n + 2 is even, because it equals 2j for an integer j = 3k +...

rectangle abcd prove: triangle adc is congruent to triangle bcd 1. Given: ABCD is a rectangle 2. AB = CD since opposite sides of rectangle are congruent 3. BC = AD since opposite sides of rectangle are congruent 4. AC = AC by the Reflexive Property of Equality 5. triangle ADC = triangle CBA by...