Marking; Direct & geometric

Marking time

Proofs Workshop

The following shows the attempts of each of three students to prove the result that

x+

1 ≥ 2 for all x > 0. x

How many marks (out of 10) would you give each of them?

Can each proof be amended so that it deserves full marks?

x+

Student 1:

1

≥ 2

x

⇒ x 2 + 1 ≥ 2x ⇒ x 2 − 2x + 1 ≥ 0 ⇒

(x − 1)2

≥ 0

Student 2:

which is always true.

1 1/x 1

Compare areas. x

Student 3:

x + 1/x x - 1/x

Pythagoras' theorem gives the proof.

2

Shirleen Stibbe

http://www.shirleenstibbe.co.uk

Marking time – Tutors' view (handout)

Proofs Workshop

1 ≥ 2 for all x > 0. x How many marks (out of 10) would you give each of them? Can each proof be amended so that it deserves full marks?

The following shows the attempts of each of three students to prove the result that x +

Student 1:

Mark 0/10

⇒

1 x+ ≥ 2 x x2 + 1 ≥ 2x

⇒

x2 − 2x +1 ≥ 0

⇒

(x − 1)2

≥ 0

which is always true

By assuming the required result, the student deduced a (known) true result and concluded from this that the required result must be true. Unfortunately it is quite possible to deduce a true result starting with a false statement. For example: 1=2 ⇒ 2 = 1 and, adding the 2 equations, ⇒ 3 = 3 which is true. But we can’t claim to have proved that 1 = 2.

To make the proof rigorous, we need to start from a true result, and deduce from it the result we require, i.e. write it ‘backwards’ as follows:

[Note: it is perfectly acceptable to start off by seeing whether the required result leads to a true statement. But then you must write it the other way up – if you can. If you can’t, as in the example of 1 = 2 above, then the required result is probably not false.]

Student 2: α

Mark 6/10

1 1/x

1 x

α

Compare areas.

This is a nice idea, but the proof needs greater care and detail. The length 1/x needs to be explained and this can be done by noting that Tan( α) = 1/x, where α is the angle between the diagonal line and the horizontal. Also the diagram is incorrect when x < 1, so we need to deal with the case when 0 < x < 1. We could argue as follows: When 0 < x < 1, we have 1/x > 1, and we could interchange the labels x and 1/x in the diagram. without changing the argument at all.

The diagram shows 2 right-angled triangles and a 1 × 1 square. In one of the triangles, the sides containing the right angle have lengths 1 and x, where x > 1, so the area is ½ × x. In the other, the lengths are 1 and 1/x, so the area is ½ × 1/x. The diagram makes it clear that the sum of these areas is greater than the area of the 1 × 1 square, i.e. ( ½ × x ) + ( ½ × 1/x ) > 1, so that x + 1/x > 2 whenever x > 1, or 0 < x < 1.

Since x + 1/x = 2 when x = 1, we conclude that x + 1/x ≥ 2 for all x > 0.

Student 3: Mark 6/10 x+1/x

x−1/x

2

Pythagoras' theorem gives the proof.

Shirleen Stibbe

This is another nice idea, but again, the devil is in the detail. The diagram doesn’t work if x < 1, since then x − 1/x is negative.

For x > 1, the student obviously noticed that applying Pythagoras to the triangle gives: (x + 1/x)2 = (x − 1/x)2 + 4, and since a square is non-negative, this leads us towards a proof.

If we ditch the diagram and use only the algebra, we can complete the proof rigorously, as follows For x > 0, (x + 1/x)2 = x2 + 2 + 1/x2 = (x2 − 2 + 1/x2) + 4 = (x − 1/x)2 + 4. therefore (x + 1/x)2 ≥ 4, since a square is non-negative and so x + 1/x ≥ 2, taking positive square roots, since x > 0.

http://www.shirleenstibbe.co.uk