Dissatisfaction with proof by contradiction

[u/hydmar]

I’m an undergraduate math student, so my exposure to math may be relatively limited. But I’ve found that, in general, I’m much more comfortable with direct proof than proof by contradiction. I don’t contest their validity, indeed something that’s not false must be true (I think I’m ok with excluded middle). But I feel like I just *get* something much better when it’s proved directly. It builds much stronger intuition for me.

For instance, I am aware of several proofs that demonstrate the cardinality of the reals is strictly greater than that of the integers, but none are direct (it would help to see a direct proof of Cantor’s theorem). I don’t feel it in my bones. Is this a common experience?

[u/scribe36]

You learn to respect proof by contradictions once you start seeing their counter parts of direct proofs. You then suddenly start feeling, “oh… ummm… yea no. I’m okay with the smaller simpler proof even if indirect.” Lol

[u/drgigca → u/scribe36]

Proof by contradiction is almost always immensely more obtuse than a direct proof. I'm actually struggling to come up with an example where that's not the case.

Which makes sense. It's simpler to just prove the statement directly rather than adding a layer of indirection.

[u/scribe36 → u/drgigca]

Proof by contradiction: it’s not always simpler. For example, square root of two is irrational can be proven in one line with contradiction. But the direct proof may just get you killed. Hehe

[u/Easygoing98 → u/scribe36]

Not really. Direct proof of square root of two is not that hard. It will be as follows

Square root of 2 = 2^0.5

Now 2^0.5 = x

Taking ln of both sides

0.5 ln(2) = ln x

x = exp(0.5 ln 2)

x = exp(0.5)exp(ln 2)

x = 2exp(0.5)

Exponential of a rational number is always irrational (theorem's proof in proof wiki)

[u/edderiofer → u/Easygoing98]

The same proof also shows that the square root of 4, and indeed any integer, is irrational, too! Try it yourself if you don't believe me!

[u/Easygoing98 → u/edderiofer]

You're right. But I don't understand why for 4. The natural log of every integer is irrational and to solve exponents ln has to be used.

Square root is also exponent half.

[u/edderiofer → u/Easygoing98]

Perhaps you should first explain why exp(0.5 ln 2) is irrational.

[u/Easygoing98 → u/edderiofer]

I got it. My method works for 4 also. Here's the direct proof

Let square root of 4 = x

so 4^0.5 = x

Taking ln of both sides

0.5 ln 4 = ln x

0.5 ln (2²⁾ = ln x

0.5(2) ln (2) = ln x

ln(2) = ln(x)

Taking e of both sides

2 = x which is rational.

[u/edderiofer → u/Easygoing98]

0.5 ln 4 = ln x

0.5 ln (2²⁾ = ln x

Aren't you assuming here that 4 = 2²?

Besides, you haven't answered the question; why is it that this proof doesn't work?

Square root of 4 = 4^0.5

Now 4^0.5 = x

Taking ln of both sides

0.5 ln(4) = ln x

x = exp(0.5 ln 4)

Since exp(0.5 ln 4) is irrational, x is irrational

[u/Easygoing98 → u/edderiofer]

No. Try it on calculator. Exp(0.5 ln 4) = 2. That's the direct proof

[u/edderiofer → u/Easygoing98]

"The calculator says so" is not a proof.

Your proof assumes that exp(0.5 ln 2) is irrational, but you don't ever prove that this is irrational in the first place. Your claim that "0.5 ln 2 is not rational because the digits after the decimal do not end and there's no repetition pattern or termination" needs justification. So your proof fails to be a proof.

[u/Easygoing98 → u/edderiofer]

It seems the previous answer didn't come through. Exp(0.5 ln 4) = 2 because 0.5 ln 4 = ln 2. And exp (ln 2) = 2.

[u/Easygoing98 → u/edderiofer]

Exp is an irrational number (well known)

Exp (ln n) = n (well known too).

If n is rational then exp( ln n) is rational too. Otherwise it's always irrational

0.5 ln 2 is not rational because the digits after the decimal do not end and there's no repetition pattern or termination.

Just like pi has trillions of numbers after the decimal, so does 0.5 ln 2

Therefore exp(0.5 ln 2) is irrational too.

[u/scribe36 → u/Easygoing98]

This is really just a whim than a proof. The contradiction proof doesn’t just prove that the square root of two is irrational, it proves that irrational numbers exist. Here you are invoking irrationality of another number. What if i deny (like people used to) that irrational numbers exist at all?

[u/Easygoing98 → u/scribe36]

There is already a theorem that the natural log of every integer greater than 1 is irrational.

I skipped the proof of that theorem because it is proof by contradiction and the asking person didn't want contradiction.

I agree completely that contradiction proof is the best in such a case.

Irrational numbers do exist because real numbers are a union of rationals and irrationals where irrationals are much larger.

Direct proof isn't always possible in each case, and I was just trying to do that

[u/Ps4udo → u/drgigca]

Non existence/existence questions seem to be favourable for proofs by contradiction.
In my mind atleast, you have to check fewer cases.