Your goal: Use the defining interval to control a floor expression.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Names you may know: greatest integer function; least integer function; fractional part.

Floor, Ceiling and Greatest Integer Functions: the key idea

The floor ⌊x⌋ is the greatest integer at most x; the ceiling ⌈x⌉ is the least integer at least x. Their definitions give ⌊x⌋≤x<⌊x⌋+1. The fractional part {x}=x−⌊x⌋ lies in [0,1), even for negative x. For an integer m, ⌊x+m⌋=⌊x⌋+m. In general floors do not distribute over addition: ⌊x+y⌋ is either ⌊x⌋+⌊y⌋ or one more. Counting multiples yields the prime exponent formula vp(n!)=∑j≥1⌊npj⌋v_p(n!)=\sum_{j\ge1}\left\lfloor\frac{n}{p^j}\right\rfloor, a finite sum once the powers exceed n.

A worked example

Find ⌊−2.3⌋, ⌈−2.3⌉ and {−2.3}.

The floor is −3, the ceiling is −2, and the fractional part is −2.3−(−3)=0.7. Truncating towards zero would give the wrong floor.

Your turn: change one thing

Find the exponent of 2 in 10!.

Try this on paper before opening the explanation.

Compare your reasoning

Count multiples of 2, then the additional factors from multiples of 4 and 8: ⌊10/2⌋+⌊10/4⌋+⌊10/8⌋=5+2+1=8.

Pause and check

A trap to avoid: Treating floor as truncation towards zero for negative inputs.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove ⌊x+y⌋ is either ⌊x⌋+⌊y⌋ or that number plus 1.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Write x=m+r,y=n+s with integers m,n and 0≤r,s<1. Then 0≤r+s<2, so its floor is 0 or 1. Therefore ⌊x+y⌋=m+n+⌊r+s⌋, which gives exactly the two possibilities.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. The floor of −2.3 is what?

Foundation

  1. −3
  2. −2
  3. 2
  4. 3
Hint

Find the greatest integer no larger than −2.3.

Answer and reasoning

−3. −3≤−2.3<−2.

2. The fractional part always lies in which interval?

Foundation

  1. (0,1]
  2. [0,1)
  3. (−1,1)
  4. [−1,0]
Hint

Subtract the floor inequality.

Answer and reasoning

[0,1). 0≤x−⌊x⌋<1.

3. What is the fractional part of −2.3?

Core

  1. 0.7
  2. −0.3
  3. 0.3
  4. −0.7
Hint

Subtract its floor −3.

Answer and reasoning

0.7. −2.3+3=0.7.

4. What is v₂(10!)?

Core

  1. 7
  2. 10
  3. 8
  4. 5
Hint

Sum floors for 2,4,8.

Answer and reasoning

8. 5+2+1=8.

5. Why can floors fail to distribute over addition?

Stretch

  1. Fractional parts can add to at least 1
  2. Integers have no floors
  3. Negative inputs are forbidden
  4. Floor always doubles
Hint

The fractional parts may create a carry.

Answer and reasoning

Fractional parts can add to at least 1. For example ⌊0.6+0.6⌋=1 while both individual floors are 0.

6. Why count multiples of p² again in vₚ(n!)?

Stretch

  1. They were omitted entirely
  2. They are all prime
  3. Every multiple has exactly two factors
  4. They contribute an additional factor of p
Hint

The first count records only one factor per multiple of p.

Answer and reasoning

They contribute an additional factor of p. Higher-power counts record the extra factors carried by those multiples.

7. The ceiling of 2.3 is what?

Foundation

  1. 0
  2. 3
  3. 2
  4. −2
Hint

Choose the least integer at least 2.3.

Answer and reasoning

3. 2<2.3≤3, so the ceiling is 3.

8. Find v₅(25!).

Core

  1. 7
  2. 6
  3. 5
  4. 25
Hint

Count multiples of 5 and 25.

Answer and reasoning

6. ⌊25/5⌋+⌊25/25⌋=5+1=6.

9. For integer m, why does ⌊x+m⌋=⌊x⌋+m?

Stretch

  1. m must be positive
  2. Adding m shifts the defining unit interval by m
  3. Floor distributes over all sums
  4. x must be an integer
Hint

Add m to ⌊x⌋≤x<⌊x⌋+1.

Answer and reasoning

Adding m shifts the defining unit interval by m. The shifted integer bounds identify the floor of x+m.

Choose your next step

Continue to Diophantine equations by factorisation. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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