proof

Prove the following statement for non-zero integers a, b, c, If a divides b and b divides c, then a divides c. If an integer a divides an integer b, then we have: b = ax for some non-zero integer x If an integer b divides an integer c, then we have: c = by for some non-zero integer y Since b...

Take two integers, r and s. We can write r as a/b for integers a and b since a rational number can be written as a quotient of integers We can write s as c/d for integers c and d since a rational number can be written as a quotient of integers Add r and s: r + s = a/b + c/d With a common...

If a is an even integer and b is an odd integer then prove a − b is an odd integer Let a be our even integer Let b be our odd integer We can express a = 2x (Standard form for even numbers) for some integer x We can express b = 2y + 1 (Standard form for odd numbers) for some integer y a - b =...

Let us take an integer x which is both even and odd. As an even integer, we write x in the form 2m for some integer m As an odd integer, we write x in the form 2n + 1 for some integer n Since both the even and odd integers are the same number, we set them equal to each other 2m = 2n + 1...

Take two consecutive integers: n, n + 1 The difference of their cubes is: (n + 1)^3 - n^3 n^3 + 3n^2 + 3n + 1 - n^3 Cancel the n^3 3n^2 + 3n + 1 Factor out a 3 from the first 2 terms: 3(n^2 + n) + 1 The first two terms are always divisible by 3 but then the + 1 makes this expression not...

Take two arbitrary integers, x and y We can express the odd integer x as 2a + 1 for some integer a We can express the odd integer y as 2b + 1 for some integer b x + y = 2a + 1 + 2b + 1 x + y = 2a + 2b + 2 Factor out a 2: x + y = 2(a + b + 1) Since 2 times any integer even or odd is always...

Take an integer n. The next alternate consecutive integer is n + 2 Subtract the difference of the squares: (n + 2)^2 - n^2 n^2 + 4n + 4 - n^2 n^2 terms cancel, we get: 4n + 4 Factor out a 4: 4(n + 1) If n is odd, n + 1 is even. 4 * even is always even If n is even, n + 1 is odd. 4 * odd is...

Prove 0! = 1 Let n be a whole number, where n! represents: The product of n and all integers below it through 1. The factorial formula for n is n! = n · (n - 1) · (n - 2) · ... · 3 · 2 · 1 Written in partially expanded form, n! is: n! = n · (n - 1)! Substitute n = 1 into this expression: n! =...

Take an integer n. The next consecutive integer is n + 1 Subtract the difference of the squares: (n + 1)^2 - n^2 n^2 + 2n + 1 - n^2 n^2 terms cancel, we get: 2n + 1 2 is even. For n, if we use an even: we have even * even = Even Add 1 we have Odd 2 is even. For n, if we use an odd: we have...

Prove P(A’) = 1 - P(A) The sample space S contains an Event A and everything not A, called A' We know P(S) = 1 P(S) = P(A U A') P(A U A') = 1 P(A) + P(A') = 1 subtract P(A) from each side: P(A’) = 1 - P(A)

Use proof by contradiction. Assume sqrt(2) is rational. This means that sqrt(2) = p/q for some integers p and q, with q <>0. We assume p and q are in lowest terms. Square both side and we get: 2 = p^2/q^2 p^2 = 2q^2 This means p^2 must be an even number which means p is also even since the...

if a divides b, then a divides bc Suppose a divides b. Then there exists an integer q such that b = aq, so that bc = a(qc) and a divides bc. Suppose that a divides c. Then there exists an integer k such that c = ak, so that bc = a(kb) and a divides bc.

If JK = PQ and PQ = ST, then JK=ST JK = PQ | Given Substitute ST for PQ since PQ = ST | Substitution JK = ST

if p=2x is even, then p^2 is also even p^2 = 2 * 2 * x^2 p^2 = 4x^2 This is true because: If x is even, then x^2 is even since two evens multiplied by each other is even and 4x^2 is even If x is odd, the x^2 is odd, but 4 times the odd number is always even since even times odd is even

n^2+n = odd Factor n^2+n: n(n + 1) We have one of two scenarios: If n is odd, then n + 1 is even. The product of an odd and even number is an even number If n is even, then n + 1 is odd. The product of an even and odd number is an even number

n^2-n = even Factor n^2-n: n(n - 1) We have one of two scenarios: If n is odd, then n - 1 is even. The product of an odd and even number is an even number If n is even, then n - 1 is odd. The product of an even and odd number is an even number

if a and b are odd then a + b is even Let a and b be positive odd integers of the form: a = 2n + 1 b = 2m + 1 a + b = 2n + 1 + 2m + 1 a + b = 2n + 2m + 1 + 1 Combing like terms, we get: a + b = 2n + 2m + 2 a + b = 2(n + m) + 2 Let k = n + m a + b = 2k + 2 Therefore a + b is even

if sc = hr and hr=ab then sc=ab sc = hr (given) Since hr = ab, we can substitute ab for hr by substitution: sc = ab

Given: 9 - 4x = -19 Prove: x = 7 Solve for x in the equation 9 - 4x = - 19 Step 1: Group constants: We need to group our constants 9 and -19. To do that, we subtract 9 from both sides -4x + 9 - 9 = -19 - 9 Step 2: Cancel 9 on the left side: -4x = -28 Step 3: Divide each side of the...

if m is odd and n is odd, then mn is odd. m = 2k +a where a = 0 or 1 n = 2l + b where b = 0 or 1 mn = (2k + a)(2l + b) = 4kl + 2kb + 2al + ab Since mn is odd, ab = 1 since a = 1 and b = 1