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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/igcse/maths/edexcel/grade-8-9-8-3-proof/flashcards","label":"Proof","subjectCode":"igcse_maths","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"4ae55e67-c6dc-4863-b9d8-e53dbdacd3cf","title":"Proof Flashcards","topicId":"7c30e247-afa6-4afc-95dd-0b0591a67267","topicSlug":"grade-8-9-8-3-proof","cards":[{"id":"bd42f64b-2f88-4265-9e52-55c1b798e4a7","cardType":"basic","points":1,"front":"What is the standard algebraic representation of an even integer?","back":"$$2n$ (where $n$ is an integer)","mcqOptions":[{"text":"$$2n$ (where $n$ is an integer)","correct":true,"explanation":"Multiplying any integer $n$ by 2 produces an integer divisible by 2, so this expression represents every even integer."},{"text":"$$2n+1$ (where $n$ is an integer)","correct":false,"explanation":"Adding 1 to twice an integer always produces an odd integer. An even integer is represented by $2n$, where $n$ is an integer."},{"text":"$$\\frac{n}{2}$ (where $n$ is an integer)","correct":false,"explanation":"Dividing an integer by 2 may produce a non-integer and does not guarantee an even result. Every even integer has the form $2n$."},{"text":"$$n+2$ (where $n$ is an integer)","correct":false,"explanation":"Adding 2 preserves whether $n$ is odd or even, so the result is not necessarily even. Multiplying an integer by 2 gives the required form $2n$."}]},{"id":"01965898-d6b8-4a9d-b5d0-d793676e7009","cardType":"basic","points":1,"front":"What are the two common algebraic forms for an odd integer?","back":"$$2n+1$ or $2n-1$","mcqOptions":[{"text":"$$2n+1$ or $2n-1$","correct":true,"explanation":"Twice an integer is even, so adding or subtracting 1 makes it odd. These forms can represent every positive or negative odd integer."},{"text":"$$2n+2$ or $2n-2$","correct":false,"explanation":"Both expressions are divisible by 2 and therefore represent even integers. Odd integers are one more or one less than an even integer: $2n+1$ or $2n-1$."},{"text":"$$n+1$ or $n-1$","correct":false,"explanation":"The parity of these expressions depends on the chosen value of $n$, so they are not always odd. Guaranteed odd integers have the forms $2n+1$ or $2n-1$."},{"text":"$$3n+1$ or $3n-1$","correct":false,"explanation":"Multiples of 3 shifted by 1 can be odd or even depending on $n$. Using twice an integer and shifting by 1 guarantees an odd result."}]},{"id":"8ec1617a-3727-40c9-b02a-b9c0afc9e44a","cardType":"basic","points":1,"front":"What algebraic expressions represent three consecutive integers?","back":"$$n$, $n+1$, $n+2$","mcqOptions":[{"text":"$$n$, $n+1$, $n+2$","correct":true,"explanation":"Consecutive integers differ by 1. Starting with an arbitrary integer $n$, the next two integers are therefore $n+1$ and $n+2$."},{"text":"$$n$, $n+2$, $n+4$","correct":false,"explanation":"These values increase by 2, so they represent three integers of the same parity rather than consecutive integers. Consecutive integers must increase by 1 each time."},{"text":"$$n-1$, $n+1$, $n+3$","correct":false,"explanation":"Although the expressions are arranged symmetrically around $n$, each adjacent pair differs by 2. The correct sequence is $n$, $n+1$, $n+2$."},{"text":"$$2n$, $2n+2$, $2n+4$","correct":false,"explanation":"These expressions represent consecutive even integers because their values differ by 2. Three consecutive integers instead have the form $n$, $n+1$, $n+2$."}]},{"id":"69c8f750-7a21-4fbd-ae09-4a80b1c74038","cardType":"basic","points":1,"front":"What algebraic expressions represent two consecutive even integers?","back":"$$2n$ and $2n+2$","mcqOptions":[{"text":"$$2n$ and $2n+2$","correct":true,"explanation":"Every even integer can be written as $2n$, and the next even integer is 2 greater. Thus consecutive even integers differ by 2."},{"text":"$$2n$ and $2n+4$","correct":false,"explanation":"The expressions differ by 4, so there is another even integer, $2n+2$, between them. Consecutive even integers must differ by 2."},{"text":"$$n$ and $n+2$","correct":false,"explanation":"The expressions differ by 2 but are not guaranteed to be even because $n$ could be odd. Starting with $2n$ guarantees both integers are even."},{"text":"$$2n+1$ and $2n+3$","correct":false,"explanation":"Both expressions are odd because 1 or 3 is added to an even number. Consecutive even integers are represented by $2n$ and $2n+2$."}]},{"id":"af1ae9c4-0c10-4086-b8a1-f8f936b07e6b","cardType":"basic","points":1,"front":"What algebraic expressions represent two consecutive odd integers?","back":"$$2n+1$ and $2n+3$","mcqOptions":[{"text":"$$2n+1$ and $2n+3$","correct":true,"explanation":"Adding 1 to the even integer $2n$ produces an odd integer, and adding a further 2 gives the next odd integer."},{"text":"$$2n$ and $2n+2$","correct":false,"explanation":"Both expressions are even, not odd, because they are divisible by 2. Consecutive odd integers have the form $2n+1$ and $2n+3$."},{"text":"$$2n+1$ and $2n+5$","correct":false,"explanation":"These are both odd, but they differ by 4, leaving another odd integer between them. Consecutive odd integers must differ by 2."},{"text":"$$n+1$ and $n+3$","correct":false,"explanation":"The expressions differ by 2, but their parity depends on $n$ and is not guaranteed to be odd. Adding 1 to $2n$ guarantees odd values."}]},{"id":"d3927153-f9ff-4ad0-b2a5-b207709bb575","cardType":"basic","points":2,"front":"Why is testing specific numbers (e.g., $n=1, 2, 3$) not considered a mathematical proof?","back":"

Testing only shows the statement is true for specific cases
Algebraic proof shows it is true for all allowed values

","mcqOptions":[{"text":"

Testing only shows the statement is true for specific cases
Algebraic proof shows it is true for all allowed values

","correct":true,"explanation":"Checking finitely many examples cannot rule out a failure at another allowed value. Algebra treats an arbitrary value and therefore establishes the statement throughout its stated domain."},{"text":"

Testing shows the statement is true for all cases
Algebraic proof shows it is true for specific allowed values

","correct":false,"explanation":"The roles have been reversed: numerical testing establishes only the cases checked, while a valid algebraic argument establishes the result for all allowed values."},{"text":"

Testing always finds a counterexample
Algebraic proof relies on chosen numerical examples

","correct":false,"explanation":"Testing can support a conjecture but does not always find a counterexample, even when one exists. Algebraic proof uses general reasoning rather than relying on selected examples."},{"text":"

Testing is invalid for positive integers
Algebraic proof avoids using integer values entirely

","correct":false,"explanation":"Positive integers are valid mathematical inputs, so their use is not the problem. The limitation is that testing finitely many inputs cannot establish a claim for every allowed value."}]},{"id":"8d9ccbc2-7a4e-4f31-9370-6f992232bcd9","cardType":"basic","points":1,"front":"What is the goal when proving an expression is a multiple of 6?","back":"Factorise the expression into the form $6(\\text{expression})$","mcqOptions":[{"text":"Factorise the expression into the form $6(\\text{expression})$","correct":true,"explanation":"Writing the result as 6 multiplied by an integer-valued expression demonstrates directly that it is divisible by 6 and hence is a multiple of 6."},{"text":"Factorise the expression into the form $3(\\text{expression})$","correct":false,"explanation":"A factor of 3 proves only that the result is divisible by 3; it may not be even. A factor of 6 is needed to establish divisibility by 6."},{"text":"Rewrite the expression into the form $6+(\\text{expression})$","correct":false,"explanation":"Adding 6 does not generally make a quantity divisible by 6, since the remaining expression may have any remainder. The result must be written as a product with factor 6."},{"text":"Rewrite the expression into the form $(\\text{expression})^6$","correct":false,"explanation":"A sixth power is not necessarily divisible by 6; for example, $5^6$ is not. The expression must instead be shown as 6 times an integer."}]},{"id":"08065be7-be31-48eb-8aab-ee53ccfbdbd6","cardType":"basic","points":1,"front":"How do you finish a proof that results in $12n$ to ensure you get the mark?","back":"State that because $n$ is an integer, $12n$ must be a multiple of 12.","mcqOptions":[{"text":"State that because $n$ is an integer, $12n$ must be a multiple of 12.","correct":true,"explanation":"Since $n$ is an integer, multiplying it by 12 produces an integer divisible by 12. This final statement explicitly links the algebraic result to the required conclusion."},{"text":"State that because $n$ is an integer, $12n$ must be a factor of 12.","correct":false,"explanation":"The factor relationship has been reversed: $12n$ is generally a multiple of 12, not a factor of it. Stating divisibility by 12 completes the proof."},{"text":"State that because $n$ is an integer, $12n$ must be a positive integer.","correct":false,"explanation":"An integer need not be positive, so that claim introduces an unjustified restriction. Positivity is irrelevant; integer $n$ is sufficient to conclude that $12n$ is a multiple of 12."},{"text":"State that because $12$ is an integer, $n$ must be a multiple of 12.","correct":false,"explanation":"The fact that 12 is an integer places no such divisibility condition on $n$. Instead, integer $n$ ensures the product $12n$ is a multiple of 12."}]},{"id":"57fe4dff-3267-4366-8f17-12c4de9954c1","cardType":"basic","points":1,"front":"What algebraic form should you reach to prove a result is a square number?","back":"$$m^2$ or $(\\text{expression})^2$","mcqOptions":[{"text":"$$m^2$ or $(\\text{expression})^2$","correct":true,"explanation":"A square number can be written as an integer multiplied by itself, so the final algebraic form must be $m^2$ or the square of an expression."},{"text":"$$m^2+1$ or $(\\text{expression})^2+1$","correct":false,"explanation":"Adding 1 means the result is one greater than a square, not necessarily a square itself. The required form is $m^2$ or $(\\text{expression})^2$."},{"text":"$$2m^2$ or $2(\\text{expression})^2$","correct":false,"explanation":"Multiplying a square by 2 does not usually produce another square number. The result must be written directly as $m^2$ or $(\\text{expression})^2$."},{"text":"$$m^3$ or $(\\text{expression})^3$","correct":false,"explanation":"A cube has power 3, whereas a square has power 2. To prove the result is a square number, reach the form $m^2$ or $(\\text{expression})^2$."}]},{"id":"244e39e0-49f9-4632-a469-5dcf2af4c516","cardType":"basic","points":1,"front":"To prove a result is \"4 more than a multiple of 8\", what target form is needed?","back":"$$8k+4$ (where $k$ is an integer)","mcqOptions":[{"text":"$$8k+4$ (where $k$ is an integer)","correct":true,"explanation":"Every multiple of 8 has the form $8k$. Adding 4 gives exactly the numbers that are 4 more than a multiple of 8."},{"text":"$$8k$ (where $k$ is an integer)","correct":false,"explanation":"This represents a multiple of 8 itself, without the required extra 4. The target form for 4 more than a multiple of 8 is $8k+4$."},{"text":"$$8k+2$ (where $k$ is an integer)","correct":false,"explanation":"Adding 2 produces numbers that are 2 more than multiples of 8, not 4 more. The correct target form is $8k+4$."},{"text":"$$4k+4$ (where $k$ is an integer)","correct":false,"explanation":"This describes numbers 4 more than a multiple of 4, which includes values that are not 4 more than multiples of 8. The required form is $8k+4$."}]},{"id":"2d355c27-2060-4176-86bd-002f38224d78","cardType":"basic","points":2,"front":"When proving a sum is a square number, why is factorising often better than expanding?","back":"It leads directly to the squared bracket form $(n+a)^2$ rather than a long quadratic.","mcqOptions":[{"text":"It leads directly to the squared bracket form $(n+a)^2$ rather than a long quadratic.","correct":true,"explanation":"Factorising can reveal two identical brackets, such as $(n+a)(n+a)$, which immediately becomes $(n+a)^2$ and proves the result is a square."},{"text":"It leads directly to the linear bracket form $(n+a)$ rather than a long quadratic.","correct":false,"explanation":"A single linear bracket is not enough to show that an expression is a square. Factorising is useful because it can reveal the repeated bracket $(n+a)^2$."},{"text":"It leads directly to the difference bracket form $(n-a)(n+a)$ rather than a long quadratic.","correct":false,"explanation":"The form $(n-a)(n+a)$ represents a difference of two squares, not necessarily a square number. The useful target is the repeated bracket $(n+a)^2$."},{"text":"It leads directly to the cubed bracket form $(n+a)^3$ rather than a long quadratic.","correct":false,"explanation":"A cubed bracket proves a cube, not a square. Factorising helps when it produces two identical factors, giving the squared form $(n+a)^2$."}]},{"id":"d2da262f-6aa8-43ef-9a16-ac80c5c3a2b2","cardType":"basic","points":1,"front":"Expand the identity for $(a-b)^2$.","back":"$$a^2 - 2ab + b^2$","mcqOptions":[{"text":"$$a^2 - 2ab + b^2$","correct":true,"explanation":"Squaring $(a-b)$ gives $a^2-ab-ab+b^2$, which combines to $a^2-2ab+b^2$."},{"text":"$$a^2 - ab + b^2$","correct":false,"explanation":"The two cross-products, $-ab$ and $-ab$, combine to $-2ab$, not $-ab$. Therefore the expansion is $a^2-2ab+b^2$."},{"text":"$$a^2 + 2ab + b^2$","correct":false,"explanation":"This is the expansion of $(a+b)^2$; the minus sign between $a$ and $b$ makes the middle term negative. The correct result is $a^2-2ab+b^2$."},{"text":"$$a^2 - 2ab - b^2$","correct":false,"explanation":"The final term comes from multiplying $-b$ by $-b$, which gives positive $b^2$. Thus the expansion is $a^2-2ab+b^2$."}]},{"id":"b73f0fd9-94d6-4a86-a595-52a3be285a6a","cardType":"basic","points":1,"front":"What is the difference of two squares identity used in proofs?","back":"$$a^2 - b^2 = (a-b)(a+b)$","mcqOptions":[{"text":"$$a^2 - b^2 = (a-b)(a+b)$","correct":true,"explanation":"Multiplying $(a-b)(a+b)$ cancels the two cross-terms, leaving $a^2-b^2$. This identity is useful for factorising a difference of two squares."},{"text":"$$a^2 + b^2 = (a-b)(a+b)$","correct":false,"explanation":"The product $(a-b)(a+b)$ gives $a^2-b^2$ because the cross-terms cancel. It does not factorise the sum $a^2+b^2$ in this way."},{"text":"$$a^2 - b^2 = (a-b)^2$","correct":false,"explanation":"Squaring $(a-b)$ introduces the extra middle term $-2ab$. The difference of two squares instead factorises into two different brackets, $(a-b)(a+b)$."},{"text":"$$a^2 - b^2 = (a+b)^2$","correct":false,"explanation":"Expanding $(a+b)^2$ gives $a^2+2ab+b^2$, including a positive middle term. The correct factorisation is $(a-b)(a+b)$."}]},{"id":"34a1e847-3eab-46be-9ee9-2bccc44c82e5","cardType":"basic","points":2,"front":"Prove that $2n+5$ is always odd for any integer $n$.","back":"

$2n+4$ is a multiple of 2 (even)
$2n+5$ is one more than an even number

","mcqOptions":[{"text":"

$2n+4$ is a multiple of 2 (even)
$2n+5$ is one more than an even number

","correct":true,"explanation":"Since $2n+4$ is divisible by 2, it is even. Adding 1 gives $2n+5$, which is one more than an even number and therefore odd."},{"text":"

$2n+4$ is a multiple of 2 (even)
$2n+5$ is one less than an even number

","correct":false,"explanation":"The expression $2n+5$ is one more than $2n+4$, not one less. Being one more than an even number proves it is odd."},{"text":"

$2n+5$ is a multiple of 2 (even)
$2n+5$ is one more than an even number

","correct":false,"explanation":"The first statement is false: $2n+5$ is not divisible by 2. Instead, $2n+4$ is even, and $2n+5$ is one more than it, so the result is odd."},{"text":"

$2n+4$ is a multiple of 2 (even)
$2n+5$ is two more than an even number

","correct":false,"explanation":"The difference between $2n+5$ and $2n+4$ is 1, not 2. The correct proof uses the fact that $2n+5$ is one more than an even number."}]},{"id":"92a82ebf-3753-44bf-aec9-7117b9f85b09","cardType":"cloze","points":null,"front":"To prove an expression is a multiple of 12, it must be written in the form $\\text{[...]}$ where $k$ is an integer.","back":"To prove an expression is a multiple of 12, it must be written in the form $\\boxed{12k}$ where $k$ is an integer.","mcqOptions":[{"text":"$$12k$","correct":true,"explanation":"An expression is a multiple of 12 when it equals 12 multiplied by an integer. Writing it as $12k$ guarantees divisibility by 12."},{"text":"$$12+k$","correct":false,"explanation":"Adding an integer to 12 does not generally produce a multiple of 12. The required form is $12k$, meaning 12 multiplied by an integer."},{"text":"$$k^{12}$","correct":false,"explanation":"Raising $k$ to the twelfth power does not ensure the result is divisible by 12, especially when $k=1$. A multiple of 12 has the form $12k$."},{"text":"$$k+12k$","correct":false,"explanation":"This simplifies to $13k$, which is a multiple of 13 rather than necessarily a multiple of 12. The correct form is $12k$."}]},{"id":"e19a2730-afcf-4984-9f7e-ec9a28628ad0","cardType":"cloze","points":null,"front":"The expansion of $(a+b)^2$ is $\\text{[...]}$.","back":"The expansion of $(a+b)^2$ is $\\boxed{a^2+2ab+b^2}$.","mcqOptions":[{"text":"$$a^2+2ab+b^2$","correct":true,"explanation":"Multiplying $(a+b)(a+b)$ gives $a^2+ab+ab+b^2$, and the two middle terms combine to $2ab$."},{"text":"$$a^2+ab+b^2$","correct":false,"explanation":"There are two cross-products, $ab$ and $ab$, so they combine to $2ab$, not $ab$. The correct expansion is $a^2+2ab+b^2$."},{"text":"$$a^2-2ab+b^2$","correct":false,"explanation":"The negative middle term comes from expanding $(a-b)^2$. With a plus sign, both cross-products are positive, giving $a^2+2ab+b^2$."},{"text":"$$a^2+2ab-b^2$","correct":false,"explanation":"The final term is $b$ multiplied by $b$, so it is positive $b^2$. Therefore $(a+b)^2$ expands to $a^2+2ab+b^2$."}]},{"id":"8cd50163-4c75-4348-996d-655043cab30d","cardType":"cloze","points":null,"front":"A square number is an expression that can be written in the form $\\text{[...]}$.","back":"A square number is an expression that can be written in the form $\\boxed{m^2}$.","mcqOptions":[{"text":"$$m^2$","correct":true,"explanation":"A square number is obtained by multiplying an expression by itself, so it has the form $m \\times m = m^2$."},{"text":"$$2m$","correct":false,"explanation":"Multiplying $m$ by 2 gives twice the expression, not its square. A square number results from multiplying $m$ by itself, giving $m^2$."},{"text":"$$m^3$","correct":false,"explanation":"The exponent 3 represents a cube because $m^3 = m \\times m \\times m$. A square number has exponent 2, so its form is $m^2$."},{"text":"$$m + 2$","correct":false,"explanation":"Adding 2 to $m$ increases it by a fixed amount but does not square it. A square number is formed by multiplying $m$ by itself, giving $m^2$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Proof","title":"Proof Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"7c30e247-afa6-4afc-95dd-0b0591a67267","walled":"$L39"}],"$L3a","$L3b"]