Substitutions in inequalities (1 Viewer)

[oompaman]

Hi,

How do substitutions in inequalities work. What's stopping me from making the sub of p=p/1+p, q=q/1+q etc into this
http://snag.gy/b2B8i.jpg

 

[panda15]

In this instance, you are substituting in the same variable. p does not equal p/1+p, so you cant use this substitution.

 

[oompaman]

But the question says that p, q and r are positive real numbers and the substitution is also positive and real... still confused

isnt it similar to this? http://snag.gy/iQIt7.jpg

 

[Trebla]

When doing substitutions you have to make sure there is no loss of the generality of the original variable.

For example, p to p/(1+p) isn't valid because p can be any positive number but p/(1+p) is constrained between 0 and 1.

This means you can't use p/(1+p) to 'represent' p.

In the other example you gave, replacing a to a^1/3 is valid because both have the same restrictions (i.e. all real numbers) and thus one can represent the other.

 

[panda15]

Say for example, p=5. With your substitution, LHS=5, RHS=5/6. p does not equal p/1+p, so you can't use this substitution. HSC inequalities are designed to be done without the use of substitution, so don't try it tomorrow.

 

[oompaman]

I think i get it now.

Thanks for your help

 

[RealiseNothing]

In this instance, you are substituting in the same variable. p does not equal p/1+p, so you cant use this substitution.

No, you can totally do that actually.

 

[RealiseNothing]

i.e. there's a past Olympiad inequalities question that can be done within 5 minutes by using a suitable substitution. They can be handy.

 

[panda15]

No, you can totally do that actually.

Wasn't aware that you could use substitution for inequalities. But the substitution OP used was incorrect. p does not equal p/p+1. They were substituting in the same variable. If they let u=p/1+p, then the substitution would be valid.

 

[RealiseNothing]

Wasn't aware that you could use substitution for inequalities. But the substitution OP used was incorrect. p does not equal p/p+1. They were substituting in the same variable. If they let u=p/1+p, then the substitution would be valid.

You can use the same variable, it's just a transformation.

 

[Trebla]

You can use the same variable, because they are all dummy variables but you have to make sure your substitutions don't change the range of the variable, as I outlined below:

When doing substitutions you have to make sure there is no loss of the generality of the original variable.

For example, p to p/(1+p) isn't valid because p can be any positive number but p/(1+p) is constrained between 0 and 1.

This means you can't use p/(1+p) to 'represent' p.

In the other example you gave, replacing a to a^1/3 is valid because both have the same restrictions (i.e. all real numbers) and thus one can represent the other.

 

[seanieg89]

The "changing of range" is not the critical issue here. This entire argument is logically misguided:

The premise is that p+q-r is non-negative. This is a statement about p,q and r. The desired conclusion is also a statement about p,q, and r.

If you were to "let" p=p+1/p (pretending for the moment that this is somehow possible...trebla has shown why this cannot be done in this particular case), etc then the premise is no longer true about the new quantities you are denoting by p,q,r. (in fact your current premise is precisely the conclusion you are aiming for, so you are proving nothing.)

 

[seanieg89]

Logically, the fallacy is exactly the same as making the following argument:

If the number n is prime, then the number n+1 is prime.