GCSE Maths: Reasoning and proof with fractions (Higher)
Part of the GCSE Maths fractions chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Reasoning and proof with fractions
The hardest exam questions ask you to show or prove a result. You must give a chain of correct steps that ends exactly at the stated result. Use algebra so that the proof works for every value, not only one example.
Key ideas
- “Show that” means every step must appear. Start from one side and reach the other.
- An example does not prove a general result. To prove something for all n, use algebra.
- To disprove a statement, one counter-example is enough.
Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.
- Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.Start from the left side: 1n − 1n + 1
- Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n and n + 1 have no common factor (they are consecutive numbers), so the lowest common denominator is their product.Common denominator = n(n + 1)
- Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1), the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.1n = 1 × (n + 1)n(n + 1)
- Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by n.1n + 1 = 1 × nn(n + 1)
- Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.1 × (n + 1) − (1 × n)n(n + 1)
- Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.Expand 1 × (n + 1):
1 × n = n
1 × 1 = 1
so 1 × (n + 1) = n + 1
1 × n = n - Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.Top = (n + 1) − n
Top = n + 1 − n - Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.Top = 1
- Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.1n(n + 1)
- State the result. The last line matches the result we were asked to show.1n − 1n + 1 = 1n(n + 1) ✓
Check. Try n = 2. The original gives 1/6. Our result gives 1/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓
Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.
Method. Rewrite every term as a difference using the result above. Then most terms cancel in pairs (a telescoping sum), leaving only the first and the last.
- Recall the result. We have just shown that one over n(n + 1) equals one over n minus one over (n + 1).1n(n + 1) = 1n − 1n + 1
- Rewrite the first terms. Put n = 1, 2, 3, … into the result.11×2 = 1 − 12
12×3 = 12 − 13
13×4 = 13 − 14 - Rewrite the last term. Put n = 9.19×10 = 19 − 110
- Add everything. Write the whole sum with the differences in brackets.(1 − 12) + (12 − 13) + (13 − 14) + … + (19 − 110)
- Cancel the pairs. The −12 cancels with the next +12, the −13 cancels with the next +13, and so on all the way along.Only 1 and −110 are left
- Work out what is left. Write 1 as tenths and subtract.1 − 110 = 1010 − 110 = 910
Check. Add the first few terms to see the pattern: 12 = 12; + 16 = 23; + 112 = 34; + 120 = 45. With n terms the total is n over (n + 1), so with 9 terms it is 910. ✓
Watch out. Write enough terms to show which ones cancel. Do not just state the answer.
Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.
- Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.Subtract the smaller from the bigger: nn + 1 − n − 1n
- Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n + 1 and n have no common factor (they are consecutive numbers), so the lowest common denominator is their product.Common denominator = n(n + 1)
- Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by n, the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.nn + 1 = n × nn(n + 1)
- Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1).n − 1n = (n − 1) × (n + 1)n(n + 1)
- Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.n × n − ((n − 1) × (n + 1))n(n + 1)
- Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.n × n = n2
Expand (n − 1) × (n + 1):
First: n × n = n2
Outer: n × 1 = n
Inner: (−1) × n = −n
Last: (−1) × 1 = −1
so (n − 1) × (n + 1) = n2 − 1 - Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.Top = n2 − (n2 − 1)
Top = n2 − n2 + 1 - Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.Top = 1
- Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.1n(n + 1)
- Decide the sign. The difference is 1 divided by n(n + 1). When n is a positive integer, both n and n + 1 are positive, so the bottom is positive, and the top is 1. A positive number divided by a positive number is positive.1n(n + 1) > 0 for every positive integer n
- State the result. The last line matches the result we were asked to show.nn + 1 > n − 1n for every positive integer n ✓
Check. Try n = 2. The original gives 1/6. Our result gives 1/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓
Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.
Method. A counter-example is one example that makes the statement false. To disprove a statement, one counter-example is enough.
- Say what the statement claims. Alex says that if fraction B has a bigger denominator than fraction A, then B must be smaller than A.Claim: bigger denominator ⇒ smaller fraction
- Choose two fractions to test the claim. The claim is true when the numerators are the same. To break it, make the numerator of the fraction with the bigger denominator much bigger.Try 14 and 58: 8 is bigger than 4
- Give both fractions the same denominator. The LCM of 4 and 8 is 8.14 = 28
- Compare them. Same denominator, so compare the numerators.58 > 28, so 58 > 14
- Write the conclusion. The fraction with the bigger denominator (58) is the bigger fraction, which breaks Alex's rule.Alex is wrong.
Check. As decimals: 58 = 0.625 and 14 = 0.25, so 58 is bigger even though 8 > 4. ✓
Watch out. A counter-example must be a specific case that you have checked. Saying “it is not always true” without an example earns no marks.
Exercise 3F Reasoning and proof
Non-calculatorMethod. Use a common denominator to add the fractions, expand and collect the top, and compare with the result.
- Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.Start from the left side: 1n + 1n + 1
- Find a common denominator. We can only add fractions with the same bottom. The bottoms n and n + 1 have no common factor (they are consecutive numbers), so the lowest common denominator is their product.Common denominator = n(n + 1)
- Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1), the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.1n = 1 × (n + 1)n(n + 1)
- Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by n.1n + 1 = 1 × nn(n + 1)
- Combine the tops over the common denominator. Now the pieces are the same size, so we add the tops and keep the bottom.1 × (n + 1) + 1 × nn(n + 1)
- Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.Expand 1 × (n + 1):
1 × n = n
1 × 1 = 1
so 1 × (n + 1) = n + 1
1 × n = n - Write the top without brackets. Adding changes no signs.Top = n + 1 + n
- Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.Top = 2n + 1
- Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.2n + 1n(n + 1)
- State the result. The last line matches the result we were asked to show.1n + 1n + 1 = 2n + 1n(n + 1) ✓
Check. Try n = 2. The original gives 5/6. Our result gives 5/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓
Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.
Method. Use the result 1/(n(n + 1)) = 1/n − 1/(n + 1) to turn each term into a difference. Most terms then cancel in pairs.
- Recall the result. Each term has the form one over n(n + 1) for n = 1, 2, 3, … 19.1n(n + 1) = 1n − 1n + 1
- Rewrite the first terms. Put n = 1, 2, 3.11 × 2 = 1 − 12
12 × 3 = 12 − 13
13 × 4 = 13 − 14 - Rewrite the last term. Put n = 19.119 × 20 = 119 − 120
- Write the whole sum in brackets. Each bracket is one of the original terms.(1 − 12) + (12 − 13) + (13 − 14) + … + (119 − 120)
- Cancel the pairs. Each negative fraction cancels with the positive fraction in the next bracket. This is called a telescoping sum.Only 1 and −120 remain
- Work out what is left. Write 1 as twentieths: 1 = 2020. Then 2020 − 120 = 1920.1 − 120 = 1920
Check. The pattern in the exam above: with 9 terms the sum was 910. With 19 terms it is 1920, which is n over (n + 1). ✓
Watch out. The sum of the first n terms is n/(n + 1). You can use this as a quick check.
Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.
- Plan the proof. In a “show that” question the answer is given, so every step must be shown. Start from one side and use correct algebra until you reach the other side.Let x = 0.18
- Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.0.18 = 0.181818…
- Let x equal the recurring decimal. Using a letter lets us do algebra on the number.x = 0.18 = 0.181818…
- Multiply by 100. 2 digits repeat, so multiplying by 102 = 100 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.100x = 18.1818…
- Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.100x − x = 18.1818… − 0.1818…
99x = 18 - Solve for x. Divide both sides by the number in front of x.x = 1899
- Simplify the fraction. The highest common factor of 18 and 99 is 9, so divide the top and the bottom by 9.1899 = 211
- State the result. The last line must match the result we were asked to show.0.18 = 211 ✓
Check. A different route: divide the top by the bottom. 2 ÷ 11 gives 0.18, which matches the original. ✓
Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.
Method. Compare the three fractions to show that one lies between the other two, then explain how the numerators and denominators were combined.
- Change to decimals. Dividing the top by the bottom gives numbers that are easy to compare.12 = 0.5
35 = 0.6
23 = 0.667 (to 3 d.p.) - Compare. Check that the middle one fits between the others.0.5 < 0.6 < 0.667, so 35 lies between 12 and 23
- Check exactly with cross-multiplication. 12 < 35 if 1 × 5 < 3 × 2. And 35 < 23 if 3 × 3 < 2 × 5.5 < 6 ✓
9 < 10 ✓ - Add the numerators and the denominators. This is the method the question refers to.1 + 22 + 3 = 35
- Explain. Adding the tops and the bottoms gives a fraction that is always between the two original fractions (it is called the mediant). It is a kind of weighted average of the two fractions.It works for any two positive fractions.
Check. Another example: 13 and 12 give 1 + 13 + 2 = 25, and 0.333 < 0.4 < 0.5. ✓
Watch out. This is not a method for adding fractions. It only finds a fraction in between.
Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.
- Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.Start from the left side: n + 1n − nn + 1
- Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n and n + 1 have no common factor (they are consecutive numbers), so the lowest common denominator is their product.Common denominator = n(n + 1)
- Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1), the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.n + 1n = (n + 1) × (n + 1)n(n + 1)
- Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by n.nn + 1 = n × nn(n + 1)
- Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.(n + 1) × (n + 1) − (n × n)n(n + 1)
- Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.Expand (n + 1) × (n + 1):
First: n × n = n2
Outer: n × 1 = n
Inner: 1 × n = n
Last: 1 × 1 = 1
so (n + 1) × (n + 1) = n2 + 2n + 1
n × n = n2 - Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.Top = (n2 + 2n + 1) − n2
Top = n2 + 2n + 1 − n2 - Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.Top = 2n + 1
- Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.2n + 1n(n + 1)
- State the result. The last line matches the result we were asked to show.n + 1n − nn + 1 = 2n + 1n(n + 1) ✓
Check. Try n = 2. The original gives 5/6. Our result gives 5/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓
Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.
Method. To disprove a statement, one counter-example is enough: one case where it does not work.
- Say what Ben claims. Ben says that adding two fractions that are both less than 1 always gives a fraction that is less than 1.Claim: proper + proper = proper
- Choose two proper fractions. Choose fractions that are close to 1, so their sum is likely to pass 1.34 and 23
- Add them. The LCM of 4 and 3 is 12.34 + 23 = 912 + 812 = 1712
- Check whether the answer is proper. A proper fraction has the top smaller than the bottom.17 is bigger than 12, so 1712 is improper: 1712 = 1512
- Write the conclusion. One counter-example is enough.Ben is wrong.
Check. As decimals: 0.75 + 0.667 = 1.417, which is more than 1. ✓
Watch out. A counter-example must be a specific case with numbers that you have checked.
Method. Choose unit fractions, with different denominators, whose sum is 1.
- Start with the biggest sensible unit fraction. Take 12. We cannot use 1 because that would be one unit fraction, not three.12
- Find what is left. Take it away from 1.1 − 12 = 12
- Split the remainder into two different unit fractions. Try 13, which is smaller than 12. Then see what is left.12 − 13 = 36 − 26 = 16
- Write the three fractions. The remainder is itself a unit fraction, so we are finished.12 + 13 + 16
- Check that they are all different. The question asks for three different unit fractions, so no denominator may be repeated.The denominators 2, 3 and 6 are all different ✓
Check. 12 + 13 + 16 = 36 + 26 + 16 = 66 = 1. ✓
Watch out. There are other answers, such as 1/2 + 1/4 + 1/4 (but that repeats a fraction, so it is not allowed here).
Method. Use the size of the fractions to narrow down the possible values of a, then find b.
- Say what we know. a and b are positive whole numbers, a is smaller than b, and the two unit fractions add to a half.1a + 1b = 12 with a < b
- Find the smallest a can be. 1a is only part of the half, so it must be less than 12. That means a must be bigger than 2.a > 2, so a ≥ 3
- Find the largest a can be. If a < b then 1a is the bigger fraction, so it is more than half of 12, which is 14. So 1a > 14, which means a < 4.a < 4, so a ≤ 3
- Find a. Both conditions give a = 3.a = 3
- Find b. Subtract 13 from 12 using sixths.1b = 36 − 26 = 16, so b = 6
Check. 13 + 16 = 26 + 16 = 36 = 12. ✓
Watch out. If a = b were allowed, then a = b = 4 would also work, but the question says a < b.