Proving an Equation: Find Positive Integer x such that x^2=2x

[dumb_student]

Can some one show me how this is done?

Prove that there exists a positive integer x such that x^2=2x.

Just from the looks, I want to say this this is not possible, hence vacuous proof. I'm sure I'm wrong.

 

[daveb]

Try setting it up so you can factor it and find the possible values of x.

 

[dumb_student]

Thanks a bunch, I don't know why I did think of 2 :).

 

[Reptillian]

It's easy to see that x=2 is such an integer. Since 2 exists, it's proved.

Aww, you figured it out before I could type my answer. :(

 

[dumb_student]

It's easy to see that x=2 is such an integer. Since 2 exists, it's proved.

Aww, you figured it out before I could type my answer. :(

Its all good :), I do have another one if you can help.

 

[dumb_student]

Use proof by contradictions to show that for any selection of 3 distinct integers between 0 and 6 that at least one of those numbers will be odd.

 

[dumb_student]

Use proof by contradictions to show that for any selection of 3 distinct integers between 0 and 6 that at least one of those numbers will be odd.

numbers between 0 - 6 : { 0,1,2,3,4,5,6}
|{ 0,1,2,3,4,5,6}| = 7
so at least one odd would be 1/7?
since we are picking 3 times
does it mean the chance is:
(1/7)(1/6)(1/5) = 1/(210)

 

[acabus]

numbers between 0 - 6 : { 0,1,2,3,4,5,6}
|{ 0,1,2,3,4,5,6}| = 7
so at least one odd would be 1/7?
since we are picking 3 times
does it mean the chance is:
(1/7)(1/6)(1/5) = 1/(210)

What?...

 

[Robert1986]

numbers between 0 - 6 : { 0,1,2,3,4,5,6}
|{ 0,1,2,3,4,5,6}| = 7
so at least one odd would be 1/7?
since we are picking 3 times
does it mean the chance is:
(1/7)(1/6)(1/5) = 1/(210)

In between 0 and 6 means {1,2,3,4,5} otherwise the statement is false as {2,4,6} is a counter example.

 

[Mark44]

Use proof by contradictions to show that for any selection of 3 distinct integers between 0 and 6 that at least one of those numbers will be odd.

In between 0 and 6 means {1,2,3,4,5} otherwise the statement is false as {2,4,6} is a counter example.

The problem is not clearly stated due to the ambiguity of what "between 0 and 6" means. Does "between" here mean "strictly between", with the endpoints not included? Or does "between" here mean that the endpoints are included?

Robert1986 is assuming that the latter meaning is the one that was intended.

 

[Gliese123]

Can some one show me how this is done?

Prove that there exists a positive integer x such that x^2=2x.

Just from the looks, I want to say this this is not possible, hence vacuous proof. I'm sure I'm wrong.

Easy (I think) ;)
If you divide the right side by 2 so it becomes x²/2 = x
Then you can put, for ex. 2
2²/2 = 2
OR am I wrong? xD

 

[Diffy]

Easy (I think) ;)
If you divide the right side by 2 so it becomes x²/2 = x
Then you can put, for ex. 2
2²/2 = 2
OR am I wrong? xD

A little backwards...

Although the right answer has been mentioned above,

Assume x != 0

(I use != to mean does not equal)

then divide both sides by x!

x² = 2x
Becomes
x²/x = 2x/x

Cancel the x's

and you get x = 2. This is a valid solution because x !=0.

What others have said is to subtract and factor
x² = 2x
Subtract 2x from both sides

x² - 2x = 2x - 2x
Simplify
x² -2x = 0

Factor the x

x(x-2) = 0

Zero Product property tells us that either x = 0 or x - 2 = 0

So we have two solutions x = 0 and x = 2. Since 0 is not POSITIVE, our answer must be x = 2.

 

[Reptillian]

numbers between 0 - 6 : { 0,1,2,3,4,5,6}
|{ 0,1,2,3,4,5,6}| = 7
so at least one odd would be 1/7?
since we are picking 3 times
does it mean the chance is:
(1/7)(1/6)(1/5) = 1/(210)

That's certainly a creative approach. Proof by contradiction, you should assume that there are no odd numbers between 0 and 6 and show that it leads to a logical contradiction like 2=6 or something.

 

[HallsofIvy]

numbers between 0 - 6 : { 0,1,2,3,4,5,6}
|{ 0,1,2,3,4,5,6}| = 7
so at least one odd would be 1/7?
since we are picking 3 times
does it mean the chance is:
(1/7)(1/6)(1/5) = 1/(210)

This is NOT a probability problem. There is a "chance" of picking numbers is relevant only if we were picking numbers at random.

Even then, your calculations are wrong. The chance of picking any one number is 1/7, not "at least one odd". If you were to choose three numbers from {0, 1, 2, 3, 4, 5, 6}, without replacement, the probability they would all be even would be (4/7)(3/6)(2/5)= 4/35. Therefore the probability of "at least one odd" would be 1- 4/35= 31/35.

But again, probability has nothing to do with this. The problem was to show that, if pick three numbers from {1, 2, 3, 4, 5}, at least one must be odd. And that is obviously true because there are only two even numbers in that set.