This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. When you multiply a nonzero rational number by an irrational number, the result is ALWAYS irrational: for example, 2√3 is irrational, and (5/2)π is irrational. Again using contradiction: if rational × irrational = rational, then irrational = rational/rational = rational (rationals closed under division), contradicting irrationality. The 'nonzero' qualifier is crucial: 0 · √2 = 0, which IS rational, so we exclude that special case! To prove 3√7 is irrational: Since 3 is a nonzero rational and √7 is irrational, assume (for contradiction) that 3√7 is rational. Then 3√7 = q for some rational q. Rearranging: √7 = q/3. Since q is rational, 3 is rational and nonzero, and rationals are closed under division by nonzero rationals, q/3 is rational. So √7 would be rational. But √7 is irrational! This contradiction proves 3√7 must be irrational. Choice B correctly uses this proof by contradiction, properly applying the closure of rationals under division to reach the necessary contradiction. Choice A backwards claims multiplying by rational gives rational (false for irrationals!); Choice C uses approximation which doesn't determine rationality; Choice D makes the false generalization that any expression with a square root is irrational (but √4 = 2 is rational!). The three key facts to remember: (1) Rational + rational = rational, always. (2) Rational + irrational = irrational, always (proven by contradiction). (3) Nonzero rational × irrational = irrational, always (same contradiction structure). These are universal rules you can rely on!

This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. Rational numbers (numbers that can be written as fractions p/q with integer p and q) are closed under addition and multiplication, meaning rational + rational always equals rational, and rational × rational always equals rational. This happens because adding or multiplying fractions gives another fraction: p/q + r/s = (ps+qr)/(qs), which is still a ratio of integers (ps+qr and qs are integers if p, q, r, s are integers). The system of rationals is 'closed'—operations don't take you outside the system! Proving rational + rational = rational: Let a and b be rational, so a = p/q and b = r/s for integers p, q, r, s (with q, s ≠ 0). Then a + b = p/q + r/s = (ps + qr)/(qs). Now: is this rational? Yes, because: (1) the numerator ps + qr is an integer (integers are closed under multiplication and addition), (2) the denominator qs is a nonzero integer (product of nonzero integers is nonzero integer). So a + b is the ratio of an integer to a nonzero integer = rational by definition. This proves closure of rationals under addition! Choice C correctly proves using closure that the sum is rational, showing the key logical step of computing the exact fraction 5/6 as a ratio of integers. Choice B gives an example but doesn't provide a proof: it uses an incorrect formula like (1+1)/(2+3)=2/5, but the actual sum is 5/6, and more importantly, it doesn't explain why ALL rational sums are rational. The question asks for reasoning or proof that works for ANY rationals, not just specific numbers. Examples illustrate, but don't prove universal statements! The three key facts to remember: (1) Rational + rational = rational, always (closure property—proven by showing p/q + r/s = (ps+qr)/(qs), still a fraction). (2) Rational + irrational = irrational, always (proven by contradiction—if sum were rational, we could rearrange to show the irrational is rational, contradiction!). (3) Nonzero rational × irrational = irrational, always (same contradiction structure as addition). These are universal rules you can rely on! Quick verification: if you claim something is rational, you should (in principle) be able to write it as p/q with integer p and q. If you claim it's irrational, you should explain why it CAN'T be written that way (often via contradiction). Don't just assert—provide reasoning! That's what mathematical understanding means: knowing not just WHAT is true, but WHY it's true.

This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. When you multiply a nonzero rational number by an irrational number, the result is ALWAYS irrational: for example, 2√3 is irrational, and (5/2)π is irrational. Again using contradiction: if rational × irrational = rational, then irrational = rational/rational = rational (rationals closed under division), contradicting irrationality. The 'nonzero' qualifier is crucial: 0 · √2 = 0, which IS rational, so we exclude that special case! Proving nonzero rational × irrational = irrational by contradiction: Let r be a nonzero rational and i be irrational. Assume r · i = q for some rational q. Rearranging: i = q/r. Since q is rational, r is rational and nonzero, and rationals are closed under division (by nonzero), q/r is rational. So i is rational. Contradiction with i being irrational! Therefore r · i must be irrational when r ≠ 0. (Note: 0 · irrational = 0, which IS rational, so we need r ≠ 0.) Choice A correctly uses proof by contradiction, showing that if r × i were rational, then i = (r × i)/r would be rational divided by nonzero rational = rational, contradicting that i is irrational. Choice B has the conclusion backwards: it claims the statement is false because 0 × i = 0 is rational, but the claim specifically states 'nonzero rational,' which excludes the case r = 0. The statement is true as written with the nonzero condition! Check the logical flow: does the reasoning actually support the conclusion? The three key facts to remember: (1) Rational + rational = rational, always (closure property—proven by showing p/q + r/s = (ps+qr)/(qs), still a fraction). (2) Rational + irrational = irrational, always (proven by contradiction—if sum were rational, we could rearrange to show the irrational is rational, contradiction!). (3) Nonzero rational × irrational = irrational, always (same contradiction structure as addition). These are universal rules you can rely on! Quick verification: if you claim something is rational, you should (in principle) be able to write it as p/q with integer p and q. If you claim it's irrational, you should explain why it CAN'T be written that way (often via contradiction). Don't just assert—provide reasoning! That's what mathematical understanding means: knowing not just WHAT is true, but WHY it's true.

This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. When you multiply a nonzero rational number by an irrational number, the result is ALWAYS irrational: for example, 2√3 is irrational, and (5/2)π is irrational. Again using contradiction: if rational × irrational = rational, then irrational = rational/rational = rational (rationals closed under division), contradicting irrationality. The 'nonzero' qualifier is crucial: 0 · √2 = 0, which IS rational, so we exclude that special case! Proving nonzero rational × irrational = irrational by contradiction: Let r be a nonzero rational and i be irrational. Assume r · i = q for some rational q. Rearranging: i = q/r. Since q is rational, r is rational and nonzero, and rationals are closed under division (by nonzero), q/r is rational. So i is rational. Contradiction with i being irrational! Therefore r · i must be irrational when r ≠ 0. (Note: 0 · irrational = 0, which IS rational, so we need r ≠ 0.) Choice A correctly uses proof by contradiction that the perimeter 4√5 is irrational, showing the key logical step that 4 (nonzero rational) times √5 (irrational) must be irrational. Choice B has the conclusion backwards: it claims the operation gives rational when actually it's irrational—the contradiction proof shows this. Check the logical flow: does the reasoning actually support the conclusion? The three key facts to remember: (1) Rational + rational = rational, always (closure property—proven by showing p/q + r/s = (ps+qr)/(qs), still a fraction). (2) Rational + irrational = irrational, always (proven by contradiction—if sum were rational, we could rearrange to show the irrational is rational, contradiction!). (3) Nonzero rational × irrational = irrational, always (same contradiction structure as addition). These are universal rules you can rely on! Quick verification: if you claim something is rational, you should (in principle) be able to write it as p/q with integer p and q. If you claim it's irrational, you should explain why it CAN'T be written that way (often via contradiction). Don't just assert—provide reasoning! That's what mathematical understanding means: knowing not just WHAT is true, but WHY it's true.

This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. When you add a rational number to an irrational number, the result is ALWAYS irrational: for example, 3 + √2 is irrational, and 1/2 + π is irrational. The reasoning uses proof by contradiction: if we assume rational + irrational = rational, then we could rearrange to get irrational = rational - rational = rational (since rationals are closed under subtraction), which contradicts the fact that the number is irrational. Choice B correctly uses proof by contradiction: it assumes r+i is rational (call it q), then shows i = q-r would be rational (since rationals are closed under subtraction), contradicting that i is irrational. Choice A has the logic backwards—it assumes the conclusion (that r+i is irrational) instead of assuming the opposite for contradiction. Proof by contradiction template: (1) Start: 'Assume [opposite of what we want to prove],' (2) Consequence: 'Then [rearrange to isolate the irrational],' (3) Use closure: 'Since rationals are closed under [operation], [the irrational] = rational,' (4) Contradiction: 'But this contradicts the fact that [number] is irrational,' (5) Conclude: 'Therefore our assumption was false, so [original statement] must be true.'

This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, applying the rules to classify multiple expressions. Let's analyze each expression using our key rules: (1) Rational + rational = rational, always (closure property). (2) Rational + irrational = irrational, always (proven by contradiction). (3) Nonzero rational × irrational = irrational, always (proven by contradiction). For expression (1): 2/5 + 3/10. Both 2/5 and 3/10 are rational (they're fractions with integer numerators and denominators). By closure of rationals under addition, their sum is rational. We can verify: 2/5 + 3/10 = 4/10 + 3/10 = 7/10, which is indeed a ratio of integers, hence rational. For expression (2): π + 1/2. Here π is irrational and 1/2 is rational. By our rule that rational + irrational = irrational, the sum π + 1/2 must be irrational. If it were rational, we could rearrange to get π = (π + 1/2) - 1/2 = rational - rational = rational, contradicting that π is irrational. For expression (3): (-3) × √7. Here -3 is a nonzero rational and √7 is irrational. By our rule that nonzero rational × irrational = irrational, the product (-3)√7 must be irrational. If it were rational, we could divide by -3 to get √7 = ((-3)√7)/(-3) = rational/rational = rational, contradicting that √7 is irrational. Choice A correctly classifies all three: (1) rational (sum of rationals), (2) irrational (rational + irrational), (3) irrational (nonzero rational × irrational), and provides the correct reasoning for each. Choice B makes multiple errors: claims fractions make repeating decimals (true, but repeating decimals are rational!), says π becomes close to 3 when adding 1/2 (nonsense), and thinks -3 'cancels' the square root (multiplication doesn't cancel irrationality). The three key facts to remember: (1) Rational + rational = rational, always. (2) Rational + irrational = irrational, always. (3) Nonzero rational × irrational = irrational, always. These are universal rules you can rely on! Apply them systematically to classify any expression involving rationals and irrationals.

This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. However, operations between two irrational numbers can produce EITHER rational or irrational results—there's no universal rule: √2 · √2 = 2 (rational!), but √2 · √3 = √6 (irrational). Similarly, √5 + (-√5) = 0 (rational!), but √2 + √3 is irrational. This is why we can only make definitive statements about rational-rational operations and rational-irrational operations, not irrational-irrational. While rational operations with irrationals are predictable, irrational × irrational is NOT always irrational: Consider √2 · √2 = (√2)² = 2, which is rational! But √2 · √3 = √6, which is irrational. Similarly, irrational + irrational varies: π + (-π) = 0 (rational!), but π + √2 is irrational. The lack of a universal rule for irrational-irrational operations is why we can't make 'always' statements about them—we need specific examples to determine the result. Choice C correctly provides sound reasoning that sometimes irrational × irrational is rational and sometimes it's irrational, showing √2·√8=4 (rational) and √2·√3=√6 (irrational) as valid examples. Choice A claims irrational × irrational always equals irrational, but this isn't true: √2·√2 = 2 (rational) is a counterexample. Also, √2·√2 ≠ √4 as stated—it equals 2. Different examples, different results! Why can't we make rules for irrational + irrational or irrational × irrational? Because those operations can go either way! Examples: √2 + √2 = 2√2 (irrational) but √3 + (2 - √3) = 2 (rational). √2 · √2 = 2 (rational) but √2 · √3 = √6 (irrational). Without special structure, we can't predict. That's why the standard only asks about rational-rational and rational-irrational operations—those have universal rules!

This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. Rational numbers (numbers that can be written as fractions p/q with integer p and q) are closed under addition and multiplication, meaning rational + rational always equals rational, and rational × rational always equals rational. Proving rational + rational = rational: Let a and b be rational, so a = p/q and b = r/s for integers p, q, r, s (with q, s ≠ 0). Then a + b = p/q + r/s = (ps + qr)/(qs). Now: is this rational? Yes, because: (1) the numerator ps + qr is an integer (integers are closed under multiplication and addition), (2) the denominator qs is a nonzero integer (product of nonzero integers is nonzero integer). Choice A correctly shows that a+b = (ps+qr)/(qs) is a ratio of integers with nonzero denominator, proving it's rational. Choice B has incorrect algebra—you can't add fractions by adding numerators and denominators separately! Closure property means 'stays in the system': the rational numbers are closed under +, -, ×, ÷ (by nonzero), meaning these operations on rationals always give rationals.

This question tests your understanding of how rational and irrational numbers behave under addition—specifically, why rational + irrational always produces an irrational result. When you add a rational number to an irrational number, the result is ALWAYS irrational: for example, 3 + √2 is irrational, and 1/2 + π is irrational. The reasoning uses proof by contradiction: if we assume rational + irrational = rational, then we could rearrange to get irrational = rational - rational = rational (since rationals are closed under subtraction), which contradicts the fact that the number is irrational. So the assumption must be wrong—rational + irrational must be irrational! Let's prove 5 + √3 is irrational by contradiction: Assume 5 + √3 is rational (call it q). Then √3 = q - 5. Since q and 5 are both rational, and rationals are closed under subtraction, q - 5 is rational. So √3 is rational. But wait—√3 is irrational (it cannot be written as p/q for integers p and q)! We have a contradiction: √3 is both rational (from our assumption) and irrational (known fact). Since we reached a contradiction, our assumption must be false. Therefore, 5 + √3 must be irrational! Choice C correctly uses proof by contradiction: assumes 5 + √3 is rational (call it q), rearranges to get √3 = q - 5, notes that q - 5 is rational (by closure of rationals under subtraction), identifies the contradiction (√3 would be both rational and irrational), and concludes 5 + √3 must be irrational. Choice A incorrectly claims 'adding any number to an irrational always stays irrational'—but this is only true when adding a rational to an irrational. Adding two irrationals can give a rational result (like √3 + (-√3) = 0). The statement needs to be precise! Proof by contradiction template: (1) Start: 'Assume 5 + √3 is rational,' (2) Consequence: 'Then √3 = (5 + √3) - 5 is rational - rational,' (3) Use closure: 'Since rationals are closed under subtraction, √3 is rational,' (4) Contradiction: 'But √3 is irrational,' (5) Conclude: 'Therefore 5 + √3 must be irrational.' This elegant structure proves the general rule: rational + irrational = irrational, always!

This question tests your understanding of how rational and irrational numbers behave under addition and multiplication—specifically, when operations on these number types produce rational versus irrational results, and why. When you add a rational number to an irrational number, the result is ALWAYS irrational: for example, 3 + √2 is irrational, and 1/2 + π is irrational. The reasoning uses proof by contradiction: if we assume rational + irrational = rational, then we could rearrange to get irrational = rational - rational = rational (since rationals are closed under subtraction), which contradicts the fact that the number is irrational. So the assumption must be wrong—rational + irrational must be irrational! Proving rational + irrational = irrational by contradiction: Let r be rational and i be irrational. Assume (for contradiction) that r + i = q for some rational q. Rearranging: i = q - r. Since q and r are both rational, and rationals are closed under subtraction, q - r is rational. So i is rational. But wait—we said i is irrational! We have a contradiction: i is both rational (from our assumption) and irrational (given). Since we reached a contradiction, our assumption must be false. Therefore, r + i cannot be rational—it must be irrational! Choice B correctly uses proof by contradiction showing that if 5+√3 were rational (call it q), then √3=q-5 would be rational (difference of rationals), contradicting that √3 is irrational. Choice A has the conclusion backwards: it claims that assuming 5+√3 is irrational leads to a contradiction, so it must be rational—but the correct reasoning shows the opposite! Check the logical flow: does the reasoning actually support the conclusion? Proof by contradiction template for these problems: (1) Start: 'Assume [opposite of what we want to prove],' (2) Consequence: 'Then [rearrange to isolate the irrational],' (3) Use closure: 'Since rationals are closed under [operation], [the irrational] = rational,' (4) Contradiction: 'But this contradicts the fact that [number] is irrational,' (5) Conclude: 'Therefore our assumption was false, so [original statement] must be true.' This structure works for proving rational + irrational and rational × irrational results!