Direct Proof in Algebra, Functions, and Sets

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| Attempts: 18 | Questions: 15 | Updated: Jan 28, 2026

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1) In a direct proof that "If a|b and b|c, then a|c" (where | means divides), which is the first appropriate statement?

Assume a|c

Assume a|b and b|c

Let b = ak and c = bm for integers k,m

Let c = an for some integer n

In a direct proof, we start by assuming the hypothesis. Here, the hypothesis is that a divides b and b divides c. Therefore, we assume a|b and b|c. Then, from these, we can write b = a k and c = b m for integers k and m, so c = a(k m), meaning a|c.

Explanation

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About This Quiz

Direct Proof In Algebra, Functions, And Sets - Quiz

Think you can turn everyday math facts into solid proofs? This quiz lets you put that to the test using direct proofs in number theory, algebra, and functions. You’ll write integers as multiples, express rational numbers as fractions, and use definitions of even, odd, and “divides” to build clean arguments.... see moreAlong the way, you’ll also see how properties of functions and inequalities follow naturally from your assumptions. By the end, you’ll see direct proof as a clear, reliable method—not something mysterious!
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2) To directly prove "If x > y > 0, then x² > y²," the first step is to:

Assume x² > y²

Write x = y + d where d > 0

Assume x > y > 0

Factor x² - y² = (x-y)(x+y)

In a direct proof, we begin by assuming the hypothesis, which is x > y > 0. Then, since x > y and both are positive, we can multiply the inequalities without sign change, so x² > x y and x y > y², thus x² > y².

Explanation

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3) In the direct proof that "If m and n are odd integers, then m+n is even," which statement correctly formalizes the premise?

Let m+n = 2p for some integer p

Assume m and n are not even

Suppose m+n is not odd

Let m = 2k+1 and n = 2l+1 for integers k,l

To prove directly, we formalize the hypothesis that m and n are odd. This means m = 2k+1 and n = 2l+1 for integers k and l. Then, m+n = 2k+1 + 2l+1 = 2(k+l+1), which is even.

Explanation

To prove directly, we formalize the hypothesis that m and n are odd. This means m = 2k+1 and n = 2l+1 for integers k and l. Then, m+n = 2k+1 + 2l+1 = 2(k+l+1), which is even.

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4) Which would be an appropriate first step in a direct proof of "If 3x+2 = 8, then x = 2"?

Assume 3x+2 = 8

Subtract 2 from both sides of 3x+2 = 8

Assume x = 2

Substitute x = 2 into 3x+2

In a direct proof, we assume the hypothesis, which is 3x+2 = 8. Then, we can algebraically solve for x: subtract 2 to get 3x=6, then divide by 3 to get x=2, thus deriving the conclusion.

Explanation

In a direct proof, we assume the hypothesis, which is 3x+2 = 8. Then, we can algebraically solve for x: subtract 2 to get 3x=6, then divide by 3 to get x=2, thus deriving the conclusion.

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5) In a direct proof that "If f and g are even functions, then f+g is even," what is the proper starting point?

Assume (f+g)(-x) = (f+g)(x) for all x

Let f and g be arbitrary functions

Assume f(-x) = f(x) and g(-x) = g(x) for all x

Suppose f+g is not even

We start by assuming the hypothesis that f and g are even functions, meaning f(-x) = f(x) and g(-x) = g(x) for all x. Then, (f+g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f+g)(x), so f+g is even.

Explanation

We start by assuming the hypothesis that f and g are even functions, meaning f(-x) = f(x) and g(-x) = g(x) for all x. Then, (f+g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f+g)(x), so f+g is even.

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6) For the direct proof of "If a and b are rational, then ab is rational," which is the correct first formal step?

Let ab = m/n where m,n are integers with n ≠ 0

Assume a and b are not irrational

Suppose ab is irrational

Let a = p/q and b = r/s where p,q,r,s are integers with q,s ≠ 0

We assume the hypothesis that a and b are rational, so we write them as fractions: a = p/q and b = r/s with integers p,q,r,s and q,s ≠ 0. Then, ab = (p r)/(q s), which is rational.

Explanation

We assume the hypothesis that a and b are rational, so we write them as fractions: a = p/q and b = r/s with integers p,q,r,s and q,s ≠ 0. Then, ab = (p r)/(q s), which is rational.

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7) If we were to prove "If n² is even, then n is even" directly, we should start by

Assuming n is even

Writing n = 2m for some integer m

Assuming n² is even

Writing n² = 2k for some integer k

In a direct proof, we assume the hypothesis that n² is even. This means n² = 2k for some integer k. Then, we need to show that n is even, which can be done by using properties of even and odd numbers.

Explanation

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8) Which is an effective first step in the direct proof that "If x < y and y < z, then x < z"?

Assume x < z

Subtract y from both inequalities

Add the inequalities x < y and y < z

Assume x < y and y < z

We start by assuming the hypothesis: x < y and y < z. Then, by the transitive property of inequality, we can conclude x < z directly.

Explanation

We start by assuming the hypothesis: x < y and y < z. Then, by the transitive property of inequality, we can conclude x < z directly.

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9) In a direct proof that "If A ⊆ B and B ⊆ C, then A ⊆ C," what should be our first assumption about an arbitrary element?

Let x ∈ C

Let x ∉ A

Let x ∉ C

Let x ∈ A

To prove A ⊆ C, we take an arbitrary element x in A. Then, since A ⊆ B, x ∈ B, and since B ⊆ C, x ∈ C. Thus, every x in A is in C, so A ⊆ C.

Explanation

To prove A ⊆ C, we take an arbitrary element x in A. Then, since A ⊆ B, x ∈ B, and since B ⊆ C, x ∈ C. Thus, every x in A is in C, so A ⊆ C.

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10) In the direct proof that "If n is divisible by 3, then n² is divisible by 9," which statement formalizes the premise correctly?

Let n² = 9m for some integer m

Assume n is not divisible by 3

Suppose n² is not divisible by 9

Let n = 3k for some integer k

We assume the hypothesis that n is divisible by 3, so n = 3k for some integer k. Then, n² = (3k)² = 9k², which is divisible by 9.

Explanation

We assume the hypothesis that n is divisible by 3, so n = 3k for some integer k. Then, n² = (3k)² = 9k², which is divisible by 9.

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11) Which approach would work for a direct proof of "If x is irrational and a non-zero r is rational, then xr is irrational"?

Assume xr is rational and derive a contradiction

Use induction on the denominator of r

Assume x is irrational and r = a/b, then show xr cannot be rational

Prove the contrapositive instead

For a direct proof, we assume the hypothesis: x is irrational and r is rational with r ≠ 0, so r = a/b for integers a,b with b≠0. Then, if xr were rational, say xr = c/d, then x = (c/d)/(a/b) = bc/(ad), which is rational, contradicting x irrational. Thus, xr cannot be rational.

Explanation

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12) In proving "If a function f is differentiable at x = c, then f is continuous at x = c" directly, we start by:

Assuming f is continuous at x = c

Using the definition of differentiability to express f(c)

Supposing f is not continuous at x = c

None of the above

We use the definition of differentiability: f'(c) exists, so lim_{h→0} [f(c+h) - f(c)]/h = f'(c). Then, we can show that lim_{h→0} [f(c+h) - f(c)] = 0, which means f is continuous at c.

Explanation

We use the definition of differentiability: f'(c) exists, so lim_{h→0} [f(c+h) - f(c)]/h = f'(c). Then, we can show that lim_{h→0} [f(c+h) - f(c)] = 0, which means f is continuous at c.

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13) Which is a valid first step for the direct proof that "If |x| < 1, then x² < 1"?

Assume x² < 1

Consider cases for x ≥ 0 and x < 0

Suppose x² ≥ 1

Since |x| < 1, we have -1 < x < 1

We start from the hypothesis |x| < 1, which implies -1 < x < 1. Then, for x in (-1,1), we have x² < 1, since squaring a number between -1 and 1 gives a number less than 1.

Explanation

We start from the hypothesis |x| < 1, which implies -1 < x < 1. Then, for x in (-1,1), we have x² < 1, since squaring a number between -1 and 1 gives a number less than 1.

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14) For the statement "If a sequence converges, then it is bounded," which approach works for a direct proof?

Assume the sequence is unbounded and derive a contradiction

Prove the contrapositive statement

Use mathematical induction on sequence terms

Use the definition of convergence to find bounds

We use the definition of convergence: if sequence a_n converges to L, then for ε=1, there exists N such that for n>N, |a_n - L|<1. Then, |a_n| < |L|+1 for n>N. For n≤N, the sequence is bounded since it's finite. Thus, the sequence is bounded.

Explanation

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