Analyzing if 'x<y ⇒ x^2<y^2': What Axiom to Use?

[garyljc]

i was wondering if I'm using the right approach for this
the question reads
is the following statement true for all x and y : 'If x<y then x^2<y^2'
then it follows by asking about 'if x^2<y^2'

i am currently using case analysis to do this
by considering whether x and y are positive or negative
is there any other more efficient way to do this ?

what axiom should i use in this case ? because squaring both sides instead of multiplying by a common factor

thanks

 

[mathman]

i am currently using case analysis to do this
by considering whether x and y are positive or negative
is there any other more efficient way to do this ?

Probably not. I assume x and y are supposed to be real numbers.

 

[garyljc]

so what would you suggest for me to prove these statements ?
induction ?

 

[garyljc]

eg x=-2 y= -1
then this statement will not hold
but how should i put in down

 

[HallsofIvy]

i was wondering if I'm using the right approach for this
the question reads
is the following statement true for all x and y : 'If x<y then x^2<y^2'
then it follows by asking about 'if x^2<y^2'

i am currently using case analysis to do this
by considering whether x and y are positive or negative
is there any other more efficient way to do this ?

what axiom should i use in this case ? because squaring both sides instead of multiplying by a common factor

thanks

For goodness sake, yes, do a "case analysis by considering whether x and y are positive or negative"!

In particular, you should consider the case x= -4 and y= -3!

Now, what do you by "if x^2< y^2"? If x^2< y^2, then what?

 

[robert Ihnot]

If x and y are both positive, we have given x<y, so x^2<xy <y^2.

 

[grmnsplx]

you have the right answer.

you disproved the statement by giving a counter example.

"eg x=-2 y= -1
then this statement will not hold
but how should i put in down"

first assume that the statement is true
if x < y for all x,y implies that x^2 < y^2, then we have that (-2 < -1 implies that) 4 < 1
but 4<1 is nonsense.

QED

 

[garyljc]

this is what i came up with for the first part

Assume that x<y is true , therefore x^n < y^n for n >0 is also true
therefore x^2 < y^2 is true
but the statement does not hold
eg x = -4 y =-3

should i put it down that way ? is that sufficient ?

 

[garyljc]

the second part says ' if x^2 > y^2 then x>y 0 ' . questions is whether it is true for all x and y

again ... didn't i just prove that the statement for part does not hold
so isn't it just copying the first part as my second part ?

 

[snipez90]

this is what i came up with for the first part

Assume that x<y is true , therefore x^n < y^n for n >0 is also true
therefore x^2 < y^2 is true
but the statement does not hold
eg x = -4 y =-3

should i put it down that way ? is that sufficient ?

Um, then you can't say x^2 < y^y is true because it ISN'T. I don't think you should say a statement is true and then give a counterexample as to why it is not.

You should consider different cases, such as when n is odd or even.

 

[garyljc]

how would then put in down ?
because i assume that the statement is true
then i recheck with the base statement

isn't that's hwo we do it ?

 

[snipez90]

Hmm ok I see how the original question was posed. Well, can you be more general? I mean you've deduced that x = -4 and y = -3 would render the statement false. What about just considering -x and -y? How does this affect the inequality. There shouldn't be that much casework.