# A-Level Maths Proof

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I need some clarification on algebraic proof for the new spec A-level maths. One question asked to prove that:

For x,y > 0 --> (x+y)/2 ≥ √xy, (The AM-GM inequality for quadratics)

Is it okay to start from the given inequality and work towards something which is true?

I did:

(x+y)^2 ≥ 4xy

x^2 - 2xy + y^2 ≥ 0

(x-y)^2 ≥ 0

Which is always true since squares are at least 0.

The problem is that the mark scheme says to work in the opposite direction. i.e.

Start with (x-y)^2 ≥ 0 and arrive at the inequality.

Is there a difference? Would I lose marks doing it the other way round?

For x,y > 0 --> (x+y)/2 ≥ √xy, (The AM-GM inequality for quadratics)

Is it okay to start from the given inequality and work towards something which is true?

I did:

(x+y)^2 ≥ 4xy

x^2 - 2xy + y^2 ≥ 0

(x-y)^2 ≥ 0

Which is always true since squares are at least 0.

The problem is that the mark scheme says to work in the opposite direction. i.e.

Start with (x-y)^2 ≥ 0 and arrive at the inequality.

Is there a difference? Would I lose marks doing it the other way round?

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I need some clarification on algebraic proof for the new spec A-level maths. One question asked to prove that:

For x,y > 0 --> (x+y)/2 ≥ √xy, (The AM-GM inequality for quadratics)

Is it okay to start from the given inequality and work towards something which is true?

I did:

(x+y)^2 ≥ 4xy

x^2 - 2xy + y^2 ≥ 0

(x-y)^2 ≥ 0

Which is always true since squares are at least 0.

The problem is that the mark scheme says to work in the opposite direction. i.e.

Start with (x-y)^2 ≥ 0 and arrive at the inequality.

Is there a difference? Would I lose marks doing it the other way round?

**3pointonefour**)I need some clarification on algebraic proof for the new spec A-level maths. One question asked to prove that:

For x,y > 0 --> (x+y)/2 ≥ √xy, (The AM-GM inequality for quadratics)

Is it okay to start from the given inequality and work towards something which is true?

I did:

(x+y)^2 ≥ 4xy

x^2 - 2xy + y^2 ≥ 0

(x-y)^2 ≥ 0

Which is always true since squares are at least 0.

The problem is that the mark scheme says to work in the opposite direction. i.e.

Start with (x-y)^2 ≥ 0 and arrive at the inequality.

Is there a difference? Would I lose marks doing it the other way round?

For A level it is safer to always work towards the thing you're trying to prove instead of away from it. But it's a very useful technique to work backwards to a true fact but only use it to help you - once you're happy you should rewrite the proof the correct way around.

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#4

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Bump

**3pointonefour**)Bump

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(Original post by

If you do that way then you have to make sure that every implication from one line to the other also works the other way. In this case they all do and you could use to show it. But this approach can lead to mistakes unless you're very confident with implications.

For A level it is safer to always work towards the thing you're trying to prove instead of away from it. But it's a very useful technique to use a true fact and work backwards but only use it to help you - once you're happy you should rewrite the proof the correct way around.

**Notnek**)If you do that way then you have to make sure that every implication from one line to the other also works the other way. In this case they all do and you could use to show it. But this approach can lead to mistakes unless you're very confident with implications.

For A level it is safer to always work towards the thing you're trying to prove instead of away from it. But it's a very useful technique to use a true fact and work backwards but only use it to help you - once you're happy you should rewrite the proof the correct way around.

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#6

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What do you mean by implications working the other way?

**3pointonefour**)What do you mean by implications working the other way?

(x+y)^2 ≥ 4xy

x^2 - 2xy + y^2 ≥ 0

What you're actually saying is, if (x+y)^2 ≥ 4xy is true then x^2 - 2xy + y^2 ≥ 0 is true.

For the proof to be correct, we need this to work the other way around i.e. if x^2 - 2xy + y^2 ≥ 0 is true then (x+y)^2 ≥ 4xy is true. Do you understand why? In this case it does work both ways.

But let's say you start from

-1 = 1

then square both sides

1 = 1

This is true so I've proved that -1 = 1, no?

The reason why this doesn't work is because the implication doesn't work the other way around i.e. 1 = 1 doesn't imply that -1 = 1 when you square root both sides.

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(Original post by

When you wrote this

(x+y)^2 ≥ 4xy

x^2 - 2xy + y^2 ≥ 0

What you're actually saying is, if (x+y)^2 ≥ 4xy is true then x^2 - 2xy + y^2 ≥ 0 is true.

For the proof to be correct, we need this to work the other way around i.e. if x^2 - 2xy + y^2 ≥ 0 is true then (x+y)^2 ≥ 4xy is true. Do you understand why? In this case it does work both ways.

But let's say you start from

-1 = 1

then square both sides

1 = 1

This is true so I've proved that -1 = 1, no?

The reason why this doesn't work is because the implication doesn't work the other way around i.e. 1 = 1 doesn't imply that -1 = 1 when you square root both sides.

**Notnek**)When you wrote this

(x+y)^2 ≥ 4xy

x^2 - 2xy + y^2 ≥ 0

What you're actually saying is, if (x+y)^2 ≥ 4xy is true then x^2 - 2xy + y^2 ≥ 0 is true.

For the proof to be correct, we need this to work the other way around i.e. if x^2 - 2xy + y^2 ≥ 0 is true then (x+y)^2 ≥ 4xy is true. Do you understand why? In this case it does work both ways.

But let's say you start from

-1 = 1

then square both sides

1 = 1

This is true so I've proved that -1 = 1, no?

The reason why this doesn't work is because the implication doesn't work the other way around i.e. 1 = 1 doesn't imply that -1 = 1 when you square root both sides.

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