How can the proof by contrapositive be used to show that x^2<x implies x<1?

  • Thread starter kathrynag
  • Start date
In summary: If you want to provide a proof using the contrapositive, that would be great!In summary, x^2>x means that x*x>x.
  • #1
kathrynag
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Homework Statement



x^2<x, then x<1

Homework Equations





The Attempt at a Solution


We will use a proof by contrapositive.
We assume x>1 and we want to show x^2>x.
Let x be greater than 1 and let x^2=x*x.
If x^2=x*x, then x*x>x. Therefore, x^2>x.
 
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  • #2
Not quite, your contrapositive is correct and your first step assuming x > 1 is correct. But then you said:

kathrynag said:
Let x be greater than 1 and let x^2=x*x.
If x^2=x*x, then x*x>x. Therefore, x^2>x.

Firstly x^2 = x*x is just taken for granted so you don't need to put it in. Secondly, why does x^2=x*x imply x*x > x ?

All we know is:
x > 1
Then multiplying both sides by x we get:
x*x > 1*x
so then
x^2 > x

We can also prove this without the contrapositive.
 
Last edited:
  • #3
kidmode01 said:
Not quite, your contrapositive is correct and your first step assuming x > 1 is correct. But then you said:



Firstly x^2 = x*x is just taken for granted so you don't need to put it in. Secondly, why does x^2=x*x imply x*x > x ?

All we know is:
x > 1
Then multiplying both sides by x we get:
x*x > 1*x
so then
x^2 > x

.

So, is this all I would have to do - the multiplying by x?
 
  • #4
Think about it for a little while. Every step has to follow logically from the previous and all we had to work with was a single assumption. Search the forum for similar inequality proofs or do a Google search on introduction to proofs. If you have a textbook with a section on proofs and the different kinds, that would also help looking at.
 
  • #5
kathrynag said:

Homework Statement



x^2<x, then x<1

Homework Equations





The Attempt at a Solution


We will use a proof by contrapositive.
We assume x>1 and we want to show x^2>x.
Let x be greater than 1 and let x^2=x*x.
If x^2=x*x, then x*x>x. Therefore, x^2>x.
As kidmode pointed out, you can do this directly. You also need to assume that x > 0, because the inequality isn't true otherwise. E.g., if x = -1/2, x^2 = 1/4 > -1/2, and if x = -2, x^2 = 4 > -2.

Start by assuming x > 0 and that x^2 < x.
Then x^2 - x < 0.
Now factor the left side and determine which values of x make it true.
 
  • #6
Whoops. I wrote it down but I didn't add it, thanks Mark44.
 
  • #7
I hate to add a complication but the negation of "x< 1" is NOT "x> 1".
 
  • #8
Whoops again.

~(x<1) = (x >= 1)

Just put a line underneath all the inequalities in your contrapositive proof lol.
You'll have to fix the negation of the right side of "if" statement.
You'll end up proving:

x^2 >= x

Thanks HallsofIvy
 
  • #9
Well, I'm supposed to prove using the contrapositive.
 
  • #10
"P implies Q" and "Not Q implies Not P" are logically equivalent. Constructing a truth table will show that the truth values of both statements are the same. So proving the statement is true or it's contrapositive is true are equivalent.
 
  • #11
kathrynag said:
Well, I'm supposed to prove using the contrapositive.
That wasn't clear to me from your first post "We will use a proof by contrapositive."

I interpreted that to mean that this was the direction you had decided to go, not one that was mandated in the problem.
 

What is the meaning of "x^2

The notation "x^2" means the square of x, or x multiplied by itself. The notation "x<1" means that x is less than 1. Therefore, "x^2

Why is it important to solve the inequality "x^2

Solving an inequality helps us understand the relationship between two values and find the range of values that satisfy the inequality. In this case, solving "x^2

What is the process for solving "x^2

To solve "x^2

Are there any special cases when solving "x^2

Yes, there are two special cases. First, if x=0, then the inequality becomes "0<0", which is not true. Therefore, x=0 is not a solution. Second, if x=1, then the inequality becomes "1<1", which is also not true. Therefore, x=1 is not a solution either.

What is the solution set for "x^2

The solution set for "x^2

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