GCSE Maths: Identities and algebraic proof (Higher)
Part of the GCSE Maths algebra basics chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Identities and algebraic proof
In Section 7.1 you met the words equation and identity. Now we use them properly.
An equation such as 2x + 1 = 9 is true only for particular values of x. (Here only x = 4 works.) You solve an equation to find those values.
An identity such as 2(x + 3) ≡ 2x + 6 is true for every value of x. Try x = 1: 8 and 8. Try x = 10: 26 and 26. The sign ≡ means “is identically equal to”. You do not solve an identity. Instead you show that both sides are the same by expanding, simplifying or factorising.
How do you tell the difference? Simplify both sides. If they turn into exactly the same expression it is an identity. If they do not, it is an equation. Trying a few numbers is not enough to prove an identity: 3x = 12 is true for x = 4 but it is an equation, not an identity. Numbers can only disprove an identity: one value where the two sides differ is enough.
Showing that two expressions are equivalent
In a “show that” question you are told the answer and you must explain how to get there. Pick the side that looks more complicated. Expand brackets, collect like terms or factorise until you reach the other side. Write every step and finish with a sentence.
Using algebra to support an argument
Algebra lets us prove statements about all numbers at once. Testing 3 + 5 = 8 and 7 + 9 = 16 shows that the sum of two odd numbers seems to be even, but it does not prove it. To prove it, use letters:
| Idea | Algebra |
|---|---|
| any integer (whole number) | n |
| consecutive integers | n, n + 1, n + 2, … |
| an even number | 2n |
| an odd number | 2n + 1 (or 2n − 1) |
| consecutive even numbers | 2n, 2n + 2, 2n + 4, … |
| consecutive odd numbers | 2n + 1, 2n + 3, … |
| a multiple of 3 | 3 × (an integer), like 3k |
If two different numbers are needed, use two different letters (m and n), because the two numbers do not have to be related. A proof always follows the same plan: (1) write the numbers using letters, (2) write the expression, (3) simplify it, (4) show it has the required form, (5) write a conclusion in words.
Disproving with a counter-example
To show that a statement is false, you only need one example where it fails. This is called a counter-example. For instance “the sum of two odd numbers is odd” is false: 3 + 5 = 8, which is even.
Key ideas
- An equation is true for particular values; an identity (≡) is true for all values.
- To show an identity, expand or factorise until both sides match. Testing numbers is not a proof.
- Even: 2n. Odd: 2n + 1. Consecutive integers: n, n + 1, n + 2. Use different letters for different numbers.
- To prove a multiple of k, show the expression can be written as k × (an integer).
- One counter-example is enough to prove a statement false.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: 3(x + 2) = 3x + 6
Right side: 3x + 6 = 3x + 6 - Compare the two sides. They are identical, so the statement is true for every value of x. That is what an identity is. (It is often written with the sign ≡.)3x + 6 and 3x + 6 are the same. ✓
- State the answer. Both sides are identical, so the statement is true for every value of x.It is an identity.
Check. Test two values: x = 2: left = 12, right = 12; x = 5: left = 21, right = 21. They match each time (this supports the answer, but the expanding above is what proves it). ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: (x + 1)2 = x2 + 2x + 1
Right side: x2 + 1 = x2 + 1 - Compare the two sides. They are different expressions, so they cannot be equal for every value of x.x2 + 2x + 1 and x2 + 1 are not the same.
- Find a value that shows they are not always equal. One value that makes the two sides different is enough to show it is not an identity.Try x = 1: left side = 4, right side = 2. These are different.
- Conclude. A statement that is true only for particular values is an equation.It is an equation: it is true only when x = 0.
Check. Put x = 0 into both sides: left = 1 and right = 1, which are equal. ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Method. Work on the more complicated side (the left). Expand, collect like terms, and factorise, to reach the right side.
- Start with the left side. It has a squared bracket, so it is the complicated side.(x + 4)2 − 16
- Write the squared bracket as two brackets. (x + 4)2 means (x + 4)(x + 4).(x + 4)2 − 16 = (x + 4)(x + 4) − 16
- Expand the brackets. Multiply every term by every term: x2, 4x, 4x and 16.(x + 4)(x + 4) − 16 = x2 + 4x + 4x + 16 − 16
- Collect like terms. 4x + 4x = 8x and 16 − 16 = 0.x2 + 4x + 4x + 16 − 16 = x2 + 8x
- Factorise. x is a common factor of x2 and 8x.x2 + 8x = x(x + 8)
- Write the conclusion. The last line must be the right side of the question.The left side simplifies to x(x + 8), which is the right side. So (x + 4)2 − 16 ≡ x(x + 8). ✓
Check. Test x = 2: left side (2 + 4)2 − 16 = 36 − 16 = 20 and right side 2 × (2 + 8) = 20. ✓ (Testing supports the answer but it is the algebra above that proves it.)
Watch out. Do not start with both sides and "do the same to both". Start with one side and change it into the other.
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Use a letter for the first integer. Let n be any integer. The next two integers are 1 more and 2 more.The three integers are n, n + 1, n + 2.
- Write the sum. Add the three numbers.n + (n + 1) + (n + 2)
- Simplify by collecting like terms. There are three ns and the numbers 1 and 2.n + (n + 1) + (n + 2) = 3n + 3
- Factorise to show a multiple of 3. A multiple of 3 is 3 times a whole number, so take out the common factor 3.3n + 3 = 3(n + 1)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.n is an integer, so n + 1 is also an integer. Therefore 3(n + 1) is 3 multiplied by an integer, so it is a multiple of 3. The sum of any three consecutive integers is a multiple of 3. ✓
Check. Test with 7, 8, 9: the sum is 24 = 3 × 8. The formula 3(n + 1) with n = 7 is 3 × 8 = 24. ✓
Watch out. "Let n be any integer" is the start of nearly every proof. Do not choose a particular number.
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write each odd number using a letter. An even number is 2 × an integer. An odd number is one more than an even number. The two odd numbers do not have to be linked, so use different letters m and n.The odd numbers are 2m + 1 and 2n + 1, where m and n are integers.
- Write the sum. (2m + 1) + (2n + 1)
- Simplify. Collect like terms: 2m + 2n + 1 + 1.(2m + 1) + (2n + 1) = 2m + 2n + 2
- Factorise to show it is even. An even number is 2 × an integer.2m + 2n + 2 = 2(m + n + 1)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.m + n + 1 is an integer, so 2(m + n + 1) is 2 multiplied by an integer. So it is even. The sum of any two odd numbers is even. ✓
Check. Test with 3 and 9: 3 + 9 = 12, which is 2 × 6. With m = 1, n = 4: 2(1 + 4 + 1) = 12. ✓
Watch out. Using the same letter for both numbers would only prove the result for equal odd numbers.
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Write the two numbers with a letter, work out the difference of their squares, and compare it with their sum.
- Write the two integers. Let n be any integer. The next integer is n + 1.The integers are n and n + 1.
- Write the difference of the squares. The bigger one minus the smaller one.(n + 1)2 − n2
- Expand (n + 1)2. (n + 1)(n + 1) = n2 + n + n + 1.(n + 1)2 − n2 = n2 + 2n + 1 − n2
- Simplify. The n2 terms cancel.n2 + 2n + 1 − n2 = 2n + 1
- Compare with the sum of the two integers. The sum is n + (n + 1).n + (n + 1) = 2n + 1
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.Both the difference of the squares and the sum of the two integers equal 2n + 1. So the difference between the squares of two consecutive integers equals their sum. ✓
Check. Test with 5 and 6: 62 − 52 = 36 − 25 = 11 and 5 + 6 = 11. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the two consecutive odd numbers. Let n be any integer. Then 2n + 1 is odd, and the next odd number is 2 more.The odd numbers are 2n + 1 and 2n + 3.
- Write the difference of the squares. The bigger square minus the smaller square.(2n + 3)2 − (2n + 1)2
- Expand each squared bracket separately. Four pieces each: (2n + 3)(2n + 3) and (2n + 1)(2n + 1).(2n + 3)2 = 4n2 + 6n + 6n + 9 = 4n2 + 12n + 9
(2n + 1)2 = 4n2 + 2n + 2n + 1 = 4n2 + 4n + 1 - Subtract, using a bracket for the second answer. The minus sign changes every sign in the second bracket.(2n + 3)2 − (2n + 1)2 = (4n2 + 12n + 9) − (4n2 + 4n + 1) = 4n2 + 12n + 9 − 4n2 − 4n − 1 = 8n + 8
- Factorise to show a multiple of 8. Take out the common factor 8.8n + 8 = 8(n + 1)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.n + 1 is an integer, so 8(n + 1) is 8 multiplied by an integer. So the difference between the squares of any two consecutive odd numbers is a multiple of 8. ✓
Check. Test with 5 and 7: 49 − 25 = 24 = 8 × 3. Here 2n + 1 = 5 so n = 2 and 8(2 + 1) = 24. ✓
Watch out. Expand each bracket separately, then subtract. Never subtract inside the bracket expansion.
Method. To show that a statement is false you need one counter-example: one value that makes it fail.
- Try some small values. Substitute n = 0, 1, 2, … into n2 + n + 41 and see if the answers are prime.n = 0: 41 (prime)
n = 1: 43 (prime)
n = 2: 47 (prime)
n = 3: 53 (prime) - Think about when it might fail. The statement is true for the first 40 values (n = 0 to 39) but a clever choice of n can break it. If n = 41 then every term has a factor of 41. If n = 40, then n + 1 = 41.n = 40: 402 + 40 + 41 = 1600 + 40 + 41 = 1681
- Check whether 1681 is prime. 1681 = 41 × 41, because 412 = 1681.41 × 41 = 1681, so 1681 is not prime.
- Conclude. One value that fails is enough.n = 40 is a counter-example, so the statement is false.
Check. Recalculate 41 × 41: 40 × 41 = 1640 and 1640 + 41 = 1681. ✓
Watch out. A counter-example must be a specific value that you have checked. Showing that the statement works for many values is never enough to prove it true.
Watch out
Checking a few numbers is not a proof. 2x = x + 4 works for x = 4, but it is an equation. In a proof, start with letters (“let n be any integer”), never with particular numbers. And make sure the last line says what has been shown, in words: “so this is a multiple of 3”.Exercise 7J Identities and proof
Non-calculatorKey words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: 3(x + 2) = 3x + 6
Right side: 3x + 6 = 3x + 6 - Compare the two sides. They are identical, so the statement is true for every value of x. That is what an identity is. (It is often written with the sign ≡.)3x + 6 and 3x + 6 are the same. ✓
- State the answer. Both sides are identical, so the statement is true for every value of x.It is an identity.
Check. Test two values: x = 2: left = 12, right = 12; x = 5: left = 21, right = 21. They match each time (this supports the answer, but the expanding above is what proves it). ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: 2x + 5 = 2x + 5
Right side: 11 = 11 - Compare the two sides. They are different expressions, so they cannot be equal for every value of x.2x + 5 and 11 are not the same.
- Find a value that shows they are not always equal. One value that makes the two sides different is enough to show it is not an identity.Try x = 0: left side = 5, right side = 11. These are different.
- Conclude. A statement that is true only for particular values is an equation.It is an equation: it is true only when x = 3.
Check. Put x = 3 into both sides: left = 11 and right = 11, which are equal. ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: x2 − 9 = x2 − 9
Right side: (x − 3)(x + 3) = x2 − 9 - Compare the two sides. They are identical, so the statement is true for every value of x. That is what an identity is. (It is often written with the sign ≡.)x2 − 9 and x2 − 9 are the same. ✓
- State the answer. Both sides are identical, so the statement is true for every value of x.It is an identity.
Check. Test two values: x = 2: left = −5, right = −5; x = 5: left = 16, right = 16. They match each time (this supports the answer, but the expanding above is what proves it). ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: 4x = 4x
Right side: x + 12 = x + 12 - Compare the two sides. They are different expressions, so they cannot be equal for every value of x.4x and x + 12 are not the same.
- Find a value that shows they are not always equal. One value that makes the two sides different is enough to show it is not an identity.Try x = 0: left side = 0, right side = 12. These are different.
- Conclude. A statement that is true only for particular values is an equation.It is an equation: it is true only when x = 4.
Check. Put x = 4 into both sides: left = 16 and right = 16, which are equal. ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: (x + 1)2 = x2 + 2x + 1
Right side: x2 + 1 = x2 + 1 - Compare the two sides. They are different expressions, so they cannot be equal for every value of x.x2 + 2x + 1 and x2 + 1 are not the same.
- Find a value that shows they are not always equal. One value that makes the two sides different is enough to show it is not an identity.Try x = 1: left side = 4, right side = 2. These are different.
- Conclude. A statement that is true only for particular values is an equation.It is an equation: it is true only when x = 0.
Check. Put x = 0 into both sides: left = 1 and right = 1, which are equal. ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: 5x − 2x = 3x
Right side: 3x = 3x - Compare the two sides. They are identical, so the statement is true for every value of x. That is what an identity is. (It is often written with the sign ≡.)3x and 3x are the same. ✓
- State the answer. Both sides are identical, so the statement is true for every value of x.It is an identity.
Check. Test two values: x = 2: left = 6, right = 6; x = 5: left = 15, right = 15. They match each time (this supports the answer, but the expanding above is what proves it). ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Method. To show that two expressions are equal, work on the more complicated one (expand the brackets and collect like terms) until it becomes the other one.
- Write the expression. There is more than one bracket. We expand each bracket separately, and then collect like terms.2(x + 5) + 3
- Expand each bracket on its own. Multiply every term in the bracket by the term in front of it, including the sign in front.Bracket: 2(x + 5)
2 × x = 2x
2 × 5 = 10
3 has no bracket, so it stays as it is. - Write all the pieces in one line. Now there are no brackets, only terms.2x + 10 + 3
- Collect the like terms. Group the terms with the same letters and powers and add their coefficients.x terms: +2x (only one of this type, so it stays as it is)
number terms: +10, +3
add the coefficients: 10 + 3 = 13 so they give 13
Answer: 2x + 13 - Compare with what we were asked to show. In a "show that" question the last line must be exactly the expression in the question, followed by a short sentence.2x + 13 is the same as 2x + 13, so the statement is shown to be true. ✓
Check. Test with numbers. Put x = 4 into the original and into your answer. The original gives 21 and the answer gives 21. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)
Watch out. Show every line of working. The last line must match the target exactly, and finish with a sentence.
Method. To show that two expressions are equal, work on the more complicated one (expand the brackets and collect like terms) until it becomes the other one.
- Write the expression. There is more than one bracket. We expand each bracket separately, and then collect like terms.4(x + 3) − 2(x + 1)
- Expand each bracket on its own. Multiply every term in the bracket by the term in front of it, including the sign in front. A minus in front of a bracket means you multiply by a negative number, so every sign inside the bracket flips.Bracket: 4(x + 3)
4 × x = 4x
4 × 3 = 12
Bracket: −2(x + 1)
(−2) × x = −2x
(−2) × 1 = −2 - Write all the pieces in one line. Now there are no brackets, only terms.4x + 12 − 2x − 2
- Collect the like terms. Group the terms with the same letters and powers and add their coefficients.x terms: +4x, −2x
add the coefficients: 4 − 2 = 2 so they give 2x
number terms: +12, −2
add the coefficients: 12 − 2 = 10 so they give 10
Answer: 2x + 10 - Compare with what we were asked to show. In a "show that" question the last line must be exactly the expression in the question, followed by a short sentence.2x + 10 is the same as 2x + 10, so the statement is shown to be true. ✓
Check. Test with numbers. Put x = 4 into the original and into your answer. The original gives 18 and the answer gives 18. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)
Watch out. Show every line of working. The last line must match the target exactly, and finish with a sentence.
Method. To show that two expressions are equal, work on the more complicated one (expand the brackets and collect like terms) until it becomes the other one.
- Write the expression. There is more than one bracket. We expand each bracket separately, and then collect like terms.x(x + 3) − 3x
- Expand each bracket on its own. Multiply every term in the bracket by the term in front of it, including the sign in front.Bracket: x(x + 3)
x × x = x2
x × 3 = 3x
−3x has no bracket, so it stays as it is. - Write all the pieces in one line. Now there are no brackets, only terms.x2 + 3x − 3x
- Collect the like terms. Group the terms with the same letters and powers and add their coefficients.x2 terms: +x2 (only one of this type, so it stays as it is)
x terms: +3x, −3x
add the coefficients: 3 − 3 = 0 so they cancel to give 0
Answer: x2 - Compare with what we were asked to show. In a "show that" question the last line must be exactly the expression in the question, followed by a short sentence.x2 is the same as x2, so the statement is shown to be true. ✓
Check. Test with numbers. Put x = 4 into the original and into your answer. The original gives 16 and the answer gives 16. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)
Watch out. Show every line of working. The last line must match the target exactly, and finish with a sentence.
Method. To show that two expressions are equal, work on the more complicated one (expand the brackets and collect like terms) until it becomes the other one.
- Write the expression. There is more than one bracket. We expand each bracket separately, and then collect like terms.3(2x − 1) − 6x
- Expand each bracket on its own. Multiply every term in the bracket by the term in front of it, including the sign in front.Bracket: 3(2x − 1)
3 × 2x = 6x
3 × (−1) = −3
−6x has no bracket, so it stays as it is. - Write all the pieces in one line. Now there are no brackets, only terms.6x − 3 − 6x
- Collect the like terms. Group the terms with the same letters and powers and add their coefficients.x terms: +6x, −6x
add the coefficients: 6 − 6 = 0 so they cancel to give 0
number terms: −3 (only one of this type, so it stays as it is)
Answer: −3 - Compare with what we were asked to show. In a "show that" question the last line must be exactly the expression in the question, followed by a short sentence.−3 is the same as −3, so the statement is shown to be true. ✓
Check. Test with numbers. Put into the original and into your answer. The original gives −3 and the answer gives −3. They are the same, so the answer is right. ✓ (One test does not prove it, but a wrong answer almost never survives it.)
Watch out. Show every line of working. The last line must match the target exactly, and finish with a sentence.
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the four numbers. Even numbers are 2 × an integer. Consecutive even numbers go up in 2s. Let n be any integer.2n, 2n + 2, 2n + 4, 2n + 6
- Write the sum. 2n + (2n + 2) + (2n + 4) + (2n + 6)
- Simplify. 4 lots of 2n make 8n. The numbers 2 + 4 + 6 make 12.2n + (2n + 2) + (2n + 4) + (2n + 6) = 8n + 12
- Factorise to show a multiple of 4. The highest common factor of 8 and 12 is 4.8n + 12 = 4(2n + 3)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.2n + 3 is an integer, so 4(2n + 3) is 4 multiplied by an integer. So the sum of any four consecutive even numbers is a multiple of 4. ✓
Check. Test with 6, 8, 10, 12: sum = 36 = 4 × 9. With 2n = 6 so n = 3: 4(2 × 3 + 3) = 36. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the two even numbers. An even number is 2 times an integer. The two numbers need different letters.2a and 2b, where a and b are integers.
- Write the sum. 2a + 2b
- Factorise. Take out the common factor 2.2a + 2b = 2(a + b)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.a + b is an integer, so 2(a + b) is 2 times an integer. So the sum of two even numbers is even. ✓
Check. Test with 4 and 10: 14 = 2 × 7, and with a = 2, b = 5: 2(2 + 5) = 14. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write an odd and an even number. Use different letters because the two numbers are not linked.Odd: 2m + 1. Even: 2n.
- Write the product. (2m + 1) × 2n
- Simplify and show a factor of 2. Multiply 2n by each term of the bracket, or just note that 2 is a factor.(2m + 1) × 2n = 2n(2m + 1)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.2n(2m + 1) has a factor of 2, so it is 2 × (the integer n(2m + 1)). So the product of an odd number and an even number is always even. ✓
Check. Test with 7 and 4: 28 is even, and 2 × 2 × (2 × 3 + 1) = 28 with m = 3, n = 2. ✓
Method. One example where the statement fails is a counter-example. Try two small odd numbers.
- Choose two odd numbers. Any two odd numbers will do, such as 3 and 5.3 and 5 are both odd.
- Add them. 3 + 5 = 8
- Decide. 8 is even, not odd.The statement says the sum is odd, but 8 is even.
- Conclude. One counter-example is enough.Chen is wrong. 3 + 5 = 8 is a counter-example.
Check. Another counter-example: 1 + 1 = 2, which is even. ✓
Watch out. A counter-example must satisfy the conditions of the statement (here: both numbers odd) but make the conclusion false.
Method. Follow the number through each instruction using algebra. Let n be the number you think of.
- Think of a number. Use a letter for the unknown starting number.n
- Double it. 2n
- Add 6. 2n + 6
- Halve it. Divide every term by 2.2n + 62 = n + 3
- Subtract the number you first thought of. Subtract n.(n + 3) − n = 3
- Conclude. The n disappeared, so the answer does not depend on the starting number.The answer is always 3.
Check. Try 10: double 20, add 6 = 26, halve = 13, subtract 10 = 3. Try 7: 14, 20, 10, 3. ✓
Watch out. The letter must disappear at the end. That is how you know the answer is the same for every starting number.
Method. Follow the number through each instruction using algebra. Let n be the starting number.
- Think of a number. n
- Add 4. n + 4
- Multiply by 3. The 3 multiplies the whole bracket.3(n + 4) = 3n + 12
- Subtract 12. 3n + 12 − 12 = 3n
- Divide by 3. 3n3 = n
- Conclude. We are back to the original number.The answer is always the number you started with.
Check. Try 5: 9, 27, 15, 5. ✓
Method. Use the meaning of even and odd: an even number is 2 × an integer, an odd number is one more than that.
- Even numbers. The even numbers are 0, 2, 4, 6, … They are 2 × 0, 2 × 1, 2 × 2, …2n is 2 times an integer, so it is even.
- Odd numbers. An odd number is next to an even number, so it is 1 more (or 1 less).2n + 1 is one more than an even number, so it is odd.
- Test with numbers. This works for any integer n.n = 3: 2n = 6 (even) and 2n + 1 = 7 (odd).
Check. n = 10: 20 is even and 21 is odd. ✓
Method. Expand the left side and compare it with the right side.
- Start with the left side. (x + 3)2 − 9
- Expand the squared bracket. (x + 3)(x + 3) = x2 + 3x + 3x + 9.(x + 3)2 − 9 = x2 + 6x + 9 − 9 = x2 + 6x
- Factorise. Take out the common factor x.x2 + 6x = x(x + 6)
- Conclude. The left side equals x(x + 6), the right side. ✓
Check. Test x = 4: (4 + 3)2 − 9 = 40 and 4 × 10 = 40. ✓
Key words. An equation can be solved: it is true only for particular values of the letter. An identity is true for every value of the letter.
Method. An identity is true for every value of the letter, an equation is true only for some values. Simplify both sides and compare.
- Simplify both sides as far as possible. If both sides turn into exactly the same expression then they are equal for every value of x. Expand any brackets and collect like terms.Left side: (x + 2)2 = x2 + 4x + 4
Right side: x2 + 4 = x2 + 4 - Compare the two sides. They are different expressions, so they cannot be equal for every value of x.x2 + 4x + 4 and x2 + 4 are not the same.
- Find a value that shows they are not always equal. One value that makes the two sides different is enough to show it is not an identity.Try x = 1: left side = 9, right side = 5. These are different.
- Conclude. A statement that is true only for particular values is an equation.It is an equation: it is true only when x = 0.
Check. Put x = 0 into both sides: left = 4 and right = 4, which are equal. ✓
Watch out. Checking one value is never enough to prove an identity: an equation can also be true for one special value.
Method. Use two different letters for the two numbers. Write each as an integer.
- Write the numbers. An odd number is 2m + 1, and an even number is 2n.2m + 1 and 2n
- Write the sum. (2m + 1) + 2n
- Rearrange to show it is odd. Take out the common factor 2 from the first and third terms.(2m + 1) + 2n = 2(m + n) + 1
- Conclude. 2(m + n) is even, and one more than an even number is odd.The sum of an odd number and an even number is always odd. ✓
Check. Test with 7 and 4: 11 is odd. ✓
Method. Expand the left side and compare it with the right side, term by term. For an identity, the x terms must match and the numbers must match.
- Expand the left side. Multiply out each bracket.3(x + a) = 3x + 3a and 2(x − 1) = 2x − 2
- Collect like terms. The x terms: 3x + 2x. The number terms: 3a and −2.5x + 3a − 2
- Match the x terms with the right side. 5x must equal bx.b = 5
- Match the number terms. 3a − 2 must equal 7. Add 2 to both sides, then divide by 3.3a = 9, so a = 3
Check. Put a = 3, b = 5 and x = 2 into both sides: 3(2 + 3) + 2(2 − 1) = 15 + 2 = 17 and 5 × 2 + 7 = 17. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the five integers. Let n be any integer.n, n + 1, n + 2, n + 3, n + 4
- Add them and simplify. Collect like terms: five ns and 1 + 2 + 3 + 4.n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 5n + 10
- Factorise. 5n + 10 = 5(n + 2)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.n + 2 is an integer, so 5(n + 2) is a multiple of 5. So the sum of any five consecutive integers is a multiple of 5. ✓
Check. Test with 3 to 7: sum 25 = 5 × 5; n = 3 gives 5(3 + 2) = 25. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the two consecutive odd numbers. Odd numbers are 2n + 1 and consecutive odd numbers differ by 2.2n + 1 and 2n + 3
- Add them and simplify. (2n + 1) + (2n + 3) = 4n + 4
- Factorise. 4n + 4 = 4(n + 1)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.n + 1 is an integer, so 4(n + 1) is a multiple of 4. So the sum of two consecutive odd numbers is always a multiple of 4. ✓
Check. Test with 9 and 11: 20 = 4 × 5; n = 4 gives 4(4 + 1) = 20. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the two consecutive integers. Let n be any integer.n and n + 1
- Write the sum of their squares. n2 + (n + 1)2
- Expand and simplify. (n + 1)2 = n2 + 2n + 1.n2 + (n + 1)2 = n2 + n2 + 2n + 1 = 2n2 + 2n + 1
- Show that it is one more than an even number. Take a factor of 2 out of the first two terms.2n2 + 2n + 1 = 2(n2 + n) + 1
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.n2 + n is an integer, so 2(n2 + n) is even, and one more than an even number is odd. So the sum of the squares of two consecutive integers is always odd. ✓
Check. Test with 4 and 5: 16 + 25 = 41, which is odd. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write two odd numbers. Use different letters, because the numbers are not linked.2m + 1 and 2n + 1
- Write the product. (2m + 1)(2n + 1)
- Expand. Four pieces: 4mn, 2m, 2n and 1.(2m + 1)(2n + 1) = 4mn + 2m + 2n + 1
- Show that it is one more than an even number. The first three terms all have a factor of 2.4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.2mn + m + n is an integer, so 2(2mn + m + n) is even, and the product is one more than an even number. So the product of two odd numbers is always odd. ✓
Check. Test with 3 and 5: 15 is odd. With m = 1, n = 2: 2(4 + 1 + 2) + 1 = 15. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the two numbers. Let n be any integer. The two numbers are 3n + 1 and 3n − 1.(3n + 1)2 − (3n − 1)2
- Expand each bracket separately. Four pieces each.(3n + 1)2 = 9n2 + 3n + 3n + 1 = 9n2 + 6n + 1
(3n − 1)2 = 9n2 − 3n − 3n + 1 = 9n2 − 6n + 1 - Subtract, with brackets. The minus changes every sign in the second bracket.(3n + 1)2 − (3n − 1)2 = (9n2 + 6n + 1) − (9n2 − 6n + 1) = 9n2 + 6n + 1 − 9n2 + 6n − 1 = 12n
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.12n is 12 multiplied by an integer, so (3n + 1)2 − (3n − 1)2 is a multiple of 12 for every integer n. ✓
Check. Test n = 2: 72 − 52 = 49 − 25 = 24 = 12 × 2. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Expand each squared bracket separately, subtract, and collect like terms.
- Write the expression. Both brackets are squares, so expand each separately.(a + b)2 − (a − b)2
- Expand each bracket separately. (a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2 - Subtract, with brackets. Every sign in the second bracket changes.(a + b)2 − (a − b)2 = (a2 + 2ab + b2) − (a2 − 2ab + b2) = 4ab
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.The left side simplifies to 4ab, which is the right side. So (a + b)2 − (a − b)2 ≡ 4ab. ✓
Check. Test a = 5, b = 2: 72 − 32 = 49 − 9 = 40 and 4 × 5 × 2 = 40. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Expand, collect like terms and factorise until you see a squared bracket.
- Write the expression. (x + 3)(x + 5) + 1
- Expand the brackets. Four pieces: x2, 5x, 3x, 15.(x + 3)(x + 5) + 1 = x2 + 5x + 3x + 15 + 1 = x2 + 8x + 16
- Factorise. Two numbers that multiply to 16 and add to 8 are 4 and 4.x2 + 8x + 16 = (x + 4)(x + 4) = (x + 4)2
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.(x + 3)(x + 5) + 1 is equal to (x + 4)2, which is a square. ✓
Check. Test x = 1: (4)(6) + 1 = 25 and (1 + 4)2 = 25. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use a letter, add the numbers, and write the answer as (a multiple of 4) plus a remainder.
- Write four consecutive integers. Let n be any integer.n, n + 1, n + 2, n + 3
- Add them and simplify. n + (n + 1) + (n + 2) + (n + 3) = 4n + 6
- Split into a multiple of 4 and a remainder. Write 6 as 4 + 2 and take out 4 from the first part.4n + 6 = 4(n + 1) + 2
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.4(n + 1) is a multiple of 4, so 4(n + 1) + 2 is 2 more than a multiple of 4. So the sum of four consecutive integers is never a multiple of 4. ✓
Check. Test with 5 to 8: 26 = 24 + 2. (Not a multiple of 4.) ✓
Method. Substitute values until one gives a non-prime number. A good idea is to try n = 11, because then every term is a multiple of 11.
- Try n = 11. 112 − 11 + 11 = 121 − 11 + 11 = 121
- Check whether 121 is prime. 121 = 11 × 11.121 is not prime.
- Conclude. n = 11 is a counter-example.The statement is false.
Check. 11 × 11 = 121. ✓
Method. Try numbers that behave differently: 0, 1 and fractions between 0 and 1.
- Try x = 1. 12 = 1, and 1 is not bigger than 1.
- Conclude. One value that fails is enough.x = 1 is a counter-example.
Check. x = 0.5: 0.52 = 0.25, which is smaller than 0.5. This is a second counter-example. ✓
Method. Try small square numbers: 1, 4, 9, 16, 25, … and add pairs.
- Add some pairs. 1 + 4 = 5, 4 + 9 = 13, 9 + 16 = 25
- Decide. 25 = 52 is a square number.32 + 42 = 52
- Conclude. 32 + 42 = 52 is a counter-example.
Check. 9 + 16 = 25 and 5 × 5 = 25. ✓
Method. Split into the two cases for n: even or odd. An integer is always one or the other.
- Write the product. Two consecutive integers are n and n + 1.n(n + 1)
- Case 1: n is even. Then n = 2k for some integer k.n(n + 1) = 2k(2k + 1), which has a factor of 2, so it is even.
- Case 2: n is odd. Then n = 2k + 1, and n + 1 = 2k + 2 = 2(k + 1).(2k + 1)(2k + 2) = 2(2k + 1)(k + 1)
- Conclude. In both cases there is a factor of 2.The product of two consecutive integers is always even. ✓
Check. Test: 6 × 7 = 42 (even), 9 × 10 = 90 (even). ✓
Watch out. This is called a proof by cases: all possibilities are covered.
Method. Expand (2n + 1)2, subtract 1, factorise, and use the fact that n(n + 1) is even.
- Expand the squared bracket. (2n + 1)(2n + 1) = 4n2 + 2n + 2n + 1.(2n + 1)2 − 1 = 4n2 + 4n + 1 − 1 = 4n2 + 4n
- Factorise. Take out the common factor 4n.4n2 + 4n = 4n(n + 1)
- Use the fact about consecutive integers. n and n + 1 are consecutive integers, so n(n + 1) is even (see the previous question). An even number is 2k.n(n + 1) = 2k for some integer k
- Substitute. 4 × 2k = 8k
- Conclude. 8k is 8 times an integer.(2n + 1)2 − 1 is a multiple of 8 for every integer n. ✓
Check. Test n = 3: 72 − 1 = 48 = 8 × 6. ✓
Key words. An integer is a whole number (it can be positive, negative or zero). Consecutive integers follow one another, like 7, 8, 9. A multiple of 3 is 3 times a whole number. To prove a statement is true for all numbers you must use algebra, because examples can only support a statement.
Method. Use letters for the numbers, write the expression, simplify it, and show that it has the form needed (for example 3 × an integer). Then write a conclusion.
- Write the expression. Both brackets are squared, so expand each separately.(n + 3)2 − (n − 2)2
- Expand each bracket separately. Four pieces each.(n + 3)2 = n2 + 3n + 3n + 9 = n2 + 6n + 9
(n − 2)2 = n2 − 2n − 2n + 4 = n2 − 4n + 4 - Subtract, with brackets. Every sign in the second bracket changes.(n + 3)2 − (n − 2)2 = (n2 + 6n + 9) − (n2 − 4n + 4) = n2 + 6n + 9 − n2 + 4n − 4 = 10n + 5
- Factorise. The common factor is 5.10n + 5 = 5(2n + 1)
- Write the conclusion. A proof must finish with a sentence in words that says what has been shown and why.5 is a factor, so the expression is a multiple of 5. And 2n + 1 is odd, and 5 × an odd number is odd. So (n + 3)2 − (n − 2)2 is always an odd multiple of 5. ✓
Check. Test n = 4: 72 − 22 = 49 − 4 = 45 = 5 × 9, an odd multiple of 5. ✓