Why?

[MSE]

Hello! I study in 7th grade.And I decided this math example. In school, teacher said me, that my solution is false. What is the problem?

Prove that: (a+3)(a-5)›(a-7)(a+5)

My Solution​

(a+3)(a-5)›(a-7)(a+5)
a2-5a+3a-15›a2+5a-7a-35. (a2=a*a)
a2-2a-15›a2-2a-35.

a2=a2;
-2a=-2a;
-15›-35;
Proven.

 

[lev888]

Hello! I study in 7th grade.And I decided this math example. In school, teacher said me, that my solution is false. What is the problem?

Prove that: (a+3)(a-5)›(a-7)(a+5)

My Solution​

(a+3)(a-5)›(a-7)(a+5)
a2-5a+3a-15›a2+5a-7a-35. (a2=a*a)
a2-2a-15›a2-2a-35.

a2=a2;
-2a=-2a;
-15›-35;
Proven.

Looks good to me. Maybe the teacher wanted you to use a particular method? E.g. move all terms to one side to simplify - you would end up with 20 > 0.

 

[JeffM]

First off, you assume what you are to prove. That is completely invalid. It is, however, a frequently useful way to find a valid proof.

Taking the work that you have done, you can see whether you can reverse the logic to get a valid proof.

[MATH]15 < 35 \implies[/MATH]
[MATH]- 15 > - 35 \implies[/MATH]
[MATH]-2a - 15 > - 2a - 35 \implies[/MATH]
[MATH]a^2 -2a - 15 > a^2 - 2a - 35 \implies[/MATH]
[MATH](a + 3)(a - 5) > (a + 5)(a - 7).[/MATH]
Do you see the difference?

Your way starts with what is to be proven, but you cannot assert it because you have not proven it yet.

The way it is done above starts with things known to be true and ends up with what is to be proven.

 

[MSE]

First off, you assume what you are to prove. That is completely invalid. It is, however, a frequently useful way to find a valid proof.

Taking the work that you have done, you can see whether you can reverse the logic to get a valid proof.

[MATH]15 < 35 \implies[/MATH]
[MATH]- 15 > - 35 \implies[/MATH]
[MATH]-2a - 15 > - 2a - 35 \implies[/MATH]
[MATH]a^2 -2a - 15 > a^2 - 2a - 35 \implies[/MATH]
[MATH](a + 3)(a - 5) > (a + 5)(a - 7).[/MATH]
Do you see the difference?

Your way starts with what is to be proven, but you cannot assert it because you have not proven it yet.

The way it is done above starts with things known to be true and ends up with what is to be proven.

Hello. I don't agree. I hypothesized that (a+3)(a-5)›(a-7)(a+5) and proved it.

 

[MSE]

Looks good to me. Maybe the teacher wanted you to use a particular method? E.g. move all terms to one side to simplify - you would end up with 20 > 0.

May be... Thanks for answer)

 

[JeffM]

Hello. I don't agree. I hypothesized that (a+3)(a-5)›(a-7)(a+5) and proved it.

Well it is not a valid proof. You cannot base a proof on hypothesizing that the proposition to be proved is true. Your teacher said you were wrong, and your teacher was correct in that assessment.

If all you wanted was someone to agree with you, why bother to ask.

 

[Dr.Peterson]

Hello. I don't agree. I hypothesized that (a+3)(a-5)›(a-7)(a+5) and proved it.

If you used words in addition to the symbols, your "proof" could become valid. Many students seem to think that words are unnecessary in a proof; in fact, they are essential. (In JeffM's version, logical symbols, which you may not have learned yet, play that role.)

You need to state how one line relates to the lines before it; as written, you have ASSUMED that the inequality is (always) true, and derived from that a true inequality. That is what you do to solve an inequality, in this case to discover that it is always true. But that is not the actual proof; what your work shows is that if the given inequality is true, then -15 > -35. Here is your work, turned into a proof by explaining the connections:

We want to prove that (a+3)(a-5)›(a-7)(a+5).​

This is equivalent to a²-5a+3a-15›a²+5a-7a-35, which is equivalent to a²-2a-15›a²-2a-35.​

​

Subtracting a²-2a from both sides, we find that this is equivalent to -15›-35, which is true independent of the variable a.​

Therefore, the original inequality is true for all a.​

Reversing the order of statements, as JeffM did, makes a neater proof.

Presumably your teacher is trying to teach you what constitutes a valid proof. Pay attention and learn!

EDIT: Here is another way to turn what you wrote into a valid proof:

We can rewrite (a+3)(a-5)›(a-7)(a+5) as the following equivalent inequality:​

a²-5a+3a-15›a²+5a-7a-35.​

a²-2a-15›a²-2a-35.​

​

But since​

a²=a²;​

-2a=-2a;​

-15›-35;​

​

we see that the equivalent inequality is always true.​

Proven.​

 

[Dr.Peterson]

Hello! I study in 7th grade.And I decided this math example. In school, teacher said me, that my solution is false. What is the problem?

Prove that: (a+3)(a-5)›(a-7)(a+5)

We've been answering your specific question, why your proof is not a proof. But it may be good to also show a rather different way to prove it. Here is what I would do:

Prove that: (a+3)(a-5) > (a-7)(a+5).​

​

This will be true if (a+3)(a-5) - (a-7)(a+5) > 0.​

​

But (a+3)(a-5) - (a-7)(a+5) = (a² - 5a + 3a - 15) - (a² + 5a - 7a - 35) = a² - 2a - 15 - a² + 2a + 35 = 20 > 0.​

​

Therefore (a+3)(a-5) > (a-7)(a+5).​

 

[MSE]

Well it is not a valid proof. You cannot base a proof on hypothesizing that the proposition to be proved is true. Your teacher said you were wrong, and your teacher was correct in that assessment.

If all you wanted was someone to agree with you, why bother to ask.

Hmm... Yes, you're right. Thanks for the reply.

 

[MSE]

We've been answering your specific question, why your proof is not a proof. But it may be good to also show a rather different way to prove it. Here is what I would do:

Prove that: (a+3)(a-5) > (a-7)(a+5).​

​

This will be true if (a+3)(a-5) - (a-7)(a+5) > 0.​

​

But (a+3)(a-5) - (a-7)(a+5) = (a² - 5a + 3a - 15) - (a² + 5a - 7a - 35) = a² - 2a - 15 - a² + 2a + 35 = 20 > 0.​

​

Therefore (a+3)(a-5) > (a-7)(a+5).​

Hmm... Yes, you're right. Thanks for the reply

 

[MSE]

If you used words in addition to the symbols, your "proof" could become valid. Many students seem to think that words are unnecessary in a proof; in fact, they are essential. (In JeffM's version, logical symbols, which you may not have learned yet, play that role.)

You need to state how one line relates to the lines before it; as written, you have ASSUMED that the inequality is (always) true, and derived from that a true inequality. That is what you do to solve an inequality, in this case to discover that it is always true. But that is not the actual proof; what your work shows is that if the given inequality is true, then -15 > -35. Here is your work, turned into a proof by explaining the connections:

We want to prove that (a+3)(a-5)›(a-7)(a+5).​

This is equivalent to a²-5a+3a-15›a²+5a-7a-35, which is equivalent to a²-2a-15›a²-2a-35.​

​

Subtracting a²-2a from both sides, we find that this is equivalent to -15›-35, which is true independent of the variable a.​

Therefore, the original inequality is true for all a.​

Reversing the order of statements, as JeffM did, makes a neater proof.

Presumably your teacher is trying to teach you what constitutes a valid proof. Pay attention and learn!

EDIT: Here is another way to turn what you wrote into a valid proof:

We can rewrite (a+3)(a-5)›(a-7)(a+5) as the following equivalent inequality:​

a²-5a+3a-15›a²+5a-7a-35.​

a²-2a-15›a²-2a-35.​

​

But since​

a²=a²;​

-2a=-2a;​

-15›-35;​

​

we see that the equivalent inequality is always true.​

Proven.​

Thank you!