# Does a mathematical proof in $\mathbb{C}$ imply a proof in the $\mathbb{R}$? [closed]

Does every proof in the complex numbers also prove the statement in the real numbers? I thought it might be true, because the real numbers are part of the complex numbers.

## closed as unclear what you're asking by Matthew Towers, quid♦Nov 18 '18 at 0:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• In the complex numbers we can prove that there is a number $x$ for which $x^2=-1$. But we can't prove this in the real numbers because it isn't true. – MJD Nov 17 '18 at 0:55
• If you prove something is true for all complex numbers in general it is true for all real numbers. But if you prove something about some complex numbers you haven't proven it about some real numbers unless you've proven it it for complex numbers with imaginary parts equal to zero. – fleablood Nov 17 '18 at 1:57
• " I thought it might be true, because the real numbers are part of the complex number." Dogs are part of animals so if you proved something about animals is it proven for dogs. Claim: some animals eat hay. (Proof:horses) So does that mean some dogs eat hay? Claim: All animals breath. (Proof: I dunno, something about cells). So does that mean all dogs breath. You are correct: real numbers are a subset of the complex. So any statement about complex will be pertainent. But not all statements are exhaustive about all possibilities. So use common sense. – fleablood Nov 17 '18 at 2:03
• As you have no doubt caught on, the fact that real numbers are a substructure of the complex numbers means all universal statements about complex numbers are true for real numbers. i.e. all statements of the form "for all complex numbers, some property (with no quantifiers) holds". Similarly, all existential statements that hold for the reals also hold in the complex numbers. So "there exists a real number $x$ such that $x^2=2$" implies "there exists a complex number $x$ such that $x^2=2.$" – spaceisdarkgreen Nov 17 '18 at 2:58
• If the statement is the sort that if it is true for the WHOLE it is true for the PART then, yes, if it's true for complex it is true for reals. But to say "all" proofs are like that is an overstatement. To make a precise statement of what is and is not implied might be more parsnickity than it seems. But whatever statements can be said about WHOLEs and PARTs can apply to reals and complex in the same way. – fleablood Nov 17 '18 at 3:51

If we show something is true for every complex number $$z\in \mathbb{C}$$, then we have shown that it is true for every $$x\in \mathbb{R}$$ since $$\mathbb{R}\subset\mathbb{C}$$.

However, not "every proof in the complex numbers" is of this form. For instance, consider the following example. We can show that there exists a $$z\in \mathbb{C}$$ such that $$z^2=-1$$, but there is no real number that has this property.

Your reasoning that if something is true about complex numbers it must be true about reals because reals are complex is sound. But I don't think you are thinking through just what sort of statements can be proven.

This is not a complete answer but you need to think about "some" and "all".

If X is true for all complex, it is true for all reals.

If Y is true for some complex, it may or may not be true for some, all, or no reals.

If W is true for no complex, it is not true for any reals.

If A is true for all reals then it is true for some complex. It may or may not be true for all conplex.

If B is true for some reals then it is true for some complex.

And if C is not true for any real it might or might not be true for some complex (but definitely not true for all).

• I think one small thing that's been overlooked is the complexity of the property. Like "x has has a square root" is true for all complex numbers and not true for all reals, in the sense of 'over the complex/real numbers'. – spaceisdarkgreen Nov 17 '18 at 3:55
• That's a good point. All complex numbers have (complex) square roots and so it is true all real numbers have (complex) square roots. But it's not true all real numbers have (real) square roots. The question is what exactly is the statement of the theorem. What doesn't happen is some magic changes like Theorem: in desserts, nuts taste good. But: in food, nuts taste bad. That's just contradictory. – fleablood Nov 17 '18 at 5:37

There is a problem in Kreyszig's Functional Analysis:

Let $$X$$ be an inner product space over $$\mathbb{C}$$, and $$T:X\to X$$ is a linear map. If $$\langle x,Tx\rangle =0\:\forall x\in X$$, then $$T$$ is the null transformation.

This is not true for inner product spaces over $$\mathbb R$$.

This depends on what you mean by a proof in the reals. Not every statement about the complex numbers is a statement about the reals. However, you can think of every complex number as a pair of real numbers (a,b), with a corresponding relationship, where one defines complex addition and multiplication as operations on ordered pairs of real numbers. So any statement you make about the complex numbers does correspond to a statement about real numbers with twice as many variables.

Other answers have already said that the validity of this statement depends on what you mean by a "proof". Here's another example of when something is true in $$\mathbb C$$ but not in $$\mathbb R$$: the fundamental theorem of algebra. It states that a polynomial of degree $$n$$ always has exactly $$n$$ roots in $$\mathbb C$$, which is however clearly not true in $$\mathbb R$$ except degenerate cases. In fact, if FTA were to be true in $$\mathbb R$$, it would be a strictly stronger statement and would imply FTA in $$\mathbb C$$, not the other way round.