Is Starting With Assumption of Truth A Valid Proof Strategy?

[Mentallic]

The question was to prove

[tex]\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}>1, x>1[/tex]

And I had two choices to go about this, I could have manipulated the expression
[tex]\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}[/tex]
by multiplying numerator and denominator by its conjugate, squaring, manipulating etc. and getting an obvious result that proves it is more than 1, but instead I went about it a quicker way which before today I thought was logically sound.I started with the assumption that it was true, and would manipulate it from there.

Squaring both sides:

[tex]x+\sqrt{x}+x-\sqrt{x}-2\sqrt{x^2-x}>1[/tex]

Rearranging:

[tex]2x-1>2\sqrt{x^2-x}[/tex]

Squaring, 2x-1>1 since x>1:

[tex]4x^2-4x+1>4x^2-4x[/tex]

[tex]1>0[/tex]

Thus since this result is true, the original statement must have been true.

I ended up getting 1/4 marks for this, and my teacher's reasoning was that it's because I started with the assumption that it was true, and any false statement can lead to a true statement. We threw counter-arguments back and forth, and after asking for another example where this happens, she gave me "if the moon is made of cheese, then cows aren't purple". Honestly, I don't get this woman.

After giving my teacher's argument about false statements leading to truth statements further thought, I admit that problems would arise if

[tex]\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}<-1[/tex]

since when I square both sides, it would lead to a truth statement. I should have proven that the original statement was at least more than -1, but anyway, I'd like to hear from you guys on what you think about my proof, her argument, and where I could improve as I honestly bear more weighting on your word than my teacher's.

 

[Mark44]

I think your argument is similar to the following, much simpler argument.

Prove that -1 >= 1

Square both sides: (-1)² >= (1)², or
1 >= 1

Since it is true that 1 is greater than or equal to itself, this apparently proves that -1 >= -1.

The problem occurs when I square both sides. Squaring is not a one-to-one operation, so there is the possibility of introducing solutions that aren't in the original solution set.

Another example that shows this idea more clearly is:
Let x = -2
Square both sides: x² = (-2)² = 4

From the first equation to the second, we have different solution sets, with {-2} for the first and {-2, 2} for the second. That means that the two equations are not equivalent.

In your problem, if you had applied only reversible operations (operations that are one-to-one), then your argument would have been valid.

 

[ojs]

If you had started by saying that it was obvious that [tex]\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} > 0[/tex] since x > 1, then I would say this proof would be good.

As for her remark about the moon, cheese and purple cows then that is a valid argument.

 

[LCKurtz]

Your argument can be made rigorous by working it, carefully, backwards:

Start with 4x² - 4x + 1 > 4x² - 4x

(2x - 1)² > 4(x² - x)

Now observe that x > 1 so both sides are positive and we can take the positive square root of both sides preserving the inequality

2x - 1 > 2 sqrt(x²-x)

We don't need absolute values at that step because 2x -1 is positive.

2x - 2 sqrt(x²-x) > 1

As you have shown, this is the same as

[tex]
\left (\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)^2 > 1
[/tex]

Now, again, since the quantity in the parentheses is clearly positive we can take the positive square root of both sides, preserving the inequality:

[tex]
\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} > 1
[/tex]

giving a perfectly valid argument.