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I come across many proofs which use just a definition of a certain concept to prove something.

E.g.: In the proof of proving the irrationality of $\sqrt{2}$, we go on assuming it to be $a/b$, where $a/b$ is rational, but is this mere definition enough for the proof?

I mean, if we were to prove something involving even numbers, we take the even number as $2n$, and not $n$ with its definition.

We represent the even number algebraically while in the previous case, we only use definitions for our proof.

So, my final question is:

Why some proofs work out only with definitions while others don't?

Thanks in advance!

mathquester
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PandaScientist
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    Every proof, if you dig sufficiently deep, is of the form "some set of definitions/assumptions" $\implies$ "some statement". – Michal Adamaszek Feb 14 '23 at 08:43
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    No; for the proof we assume that $\sqrt 2$ exists. We want to prove that it is not rational, and thus we assume that it is rational i.e. that there are two integers $a,b$ such that... – Mauro ALLEGRANZA Feb 14 '23 at 08:46
  • in that particular proof you assumed $\sqrt{2}$ to be a rational number only to later conclude that it leads to contradiction, hence $\sqrt{2}$ cannot be a rational number. – hteica Feb 14 '23 at 08:48
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    And also , we use that we can assume that $a$ and $b$ are coprime which is crucial for the final contradiction.. To be honest , I do not understand what exactly is the question. – Peter Feb 14 '23 at 08:50
  • @MichalAdamaszek, but I want to know why some proofs require just definitions and assumptions whereas others(like I mentioned) require some sort of algebraic treatment(like we define an even number as $2n$. – PandaScientist Feb 14 '23 at 08:51
  • It is a trivial fact that an even number $a$ can be written as $a=2n$ with integer $n$ because an even number is divisible by $2$ by definition. – Peter Feb 14 '23 at 08:54
  • @Peter, I get your point but I'm not asking that. I'm trying to ask why such algebraic treatment is not given to fractions in lowest terms. We just state that $a/b$ is in lowest terms but does mere stating that gives us the desired result? – PandaScientist Feb 14 '23 at 08:59
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    A slight remark : $a$ and $b$ could also be negative, but in this case we can multiply them with $-1$ to get a positive representation. And (as pointed out below) , we can then divide $a$ and $b$ by $gcd(a,b)$. Then, we have the desired representation of positive coprime integers we need. If $\sqrt{2}$ is rational, such a representation must exist because it exists for every positive rational number. And then , we show that $a$ and $b$ must be even and are done. – Peter Feb 14 '23 at 09:08
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    Not sure what the question is. Of course the definition of the terms involved matter.. But it is sometimes hard to distinguish a definition from a theorem. Here, for instance, it is a definition that a rational number is the quotient of two integers, but it is a theorem that such a quotient can be written in least terms. That is certainly a statement that requires a proof...you can't just declare it to be true. Is that what you are asking? – lulu Feb 14 '23 at 09:18
  • See, for example, this question on exactly that point. – lulu Feb 14 '23 at 09:19
  • @PandaScientist We do not just state that $a/b$ is in lowest terms. We know, from the properties of rational numbers, that for each rational number $x$, there exists some pair of coprime inegers $a,b$ such that $x=a/b$. We know this because we can prove it. – 5xum Feb 14 '23 at 09:29
  • @lulu This statement is so trivial that we do not actually need to prove it, unless we want to completely formalize the proof. – Peter Feb 14 '23 at 09:51

2 Answers2

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The proof does not define $\sqrt{2}$ as a rational number. It assumes it to be a rational number. And the rationals are defined by $\mathbb Q := \{ \frac{a}{b}, a\in\mathbb Z, b\in\mathbb N \setminus \{0\} \}$ (or some other equivalent definitions). So the order is

  1. Assume $\sqrt{2}$ was rational.

  2. Then there are $a,b$ such that $\sqrt{2} = \frac{a}{b}$ (by definition of the rationals)

  3. Conclude that this is a contradiction.

The third point is not a contradiction against the definition. It's a contradiciton against the assumption $(1)$.

mathquester
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  • I agree it assumes $\sqrt{2}$ to be rational but we also use the definition of rational numbers later (for the contradiction) that $a$ and $b$ have to be coprime. So it involved defining terms,. – PandaScientist Feb 14 '23 at 08:48
  • @PandaScientist You may use the attribute of $a,b$ being coprimes without loss of generality. Because if you wouldn't, it would just add an extra step. Imagine $a,b$ not being coprime. Then shorten the fracture by $gcd(a,b)$. Now you have coprimes and can continue. The proof will be the same. – mathquester Feb 14 '23 at 08:52
  • So what you're trying to say that if we don't assume them to be coprime, then it will be an endless cycle of proofs with the same result? – PandaScientist Feb 14 '23 at 09:01
  • I have pointed out above that the coprime property is crucial, otherwise we would not arrive at a contradiction. – Peter Feb 14 '23 at 09:11
  • @PandaScientist No, you will end up in exact the same proof. Assume $\sqrt{2}=\frac{a}{b}$, but $a,b$ are not coprime. Let $g = gcd(a,b)$. Then $\frac{a}{b} = \frac{g\cdot a'}{g\cdot b'} = \frac{a'}{b'}$. Then $\sqrt{2} = \frac{a'}{b'}$. As you see, $\sqrt{2}$ is still a rational as assumed. Being not coprime doesn't change this property. Now you are at the point where the "old" proof starts. Nice thing about proof by contradiciton is that you just need to find one contradiction in a variety of a lot of possible ways to there. – mathquester Feb 14 '23 at 09:20
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The proof assumes that $\sqrt{2}$ is a rational number, and from both that assumption and other known facts, concludes a contradiction.

The proof does not define a single thing. It just uses definitions that already exist, for example, the fact that every rational number can be written as a ratio of two integers. It then uses properties of integers to conclude that there must exist two coprime integers whose ratio equals $\sqrt 2$.

In other words:

  1. Assume $\sqrt 2 \in \mathbb Q$.
  2. It is known (i.e., a proven statement following from the definitions of integers and rational numbers) that if $x\in\mathbb Q$, then there exist two coprime numbers $a,b\in\mathbb Z$ such that $x=\frac ab$.
  3. From 1 and 2, we conclude there exist coprime $a,b\in\mathbb Z$ such that $\frac ab = \sqrt 2$.
  4. From 3, we can infer that $a^2 = 2b^2$
  5. From 4, we infer that $a$ is divisible by $2$.
  6. From 4 and 5, we infer that $b$ is divisible by $2$.
  7. 5 and 6 mean $a,b$ are not coprime, which is a contradiction, end of proof.

Where in the proof above do you see the proof "defining" anything? Nothing is defined. Every statement above is one of the following:

  • an assumption
  • a theorem (i.e., an already proven statement)
  • a consequence of previous statements following the rules of logic.
5xum
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  • I think I'm getting it now. One last question: Can you give example of a proof which involves defining something...? I mean, the proof should entirely rely upon definitions and facts derived from previous statements(kind of like a verbal proof) – PandaScientist Feb 14 '23 at 09:59
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    @PandaScientist Every proof entirely relies upon definitions and facts derived from previous statements, so I don't understand what you are asking for. – 5xum Feb 14 '23 at 10:03