HN Explorer

Ask HN:properties of real numbers

2 points by pencil ↗ HN
Hello, The addition property of real numbers says if a=b,c=d then a+c=b+d.can someone tell me why is it true? is there an algebraic proof for this or should we accept this as being true based on inductive reasoning?i really tried doing a google search for a proof but couldn't find any. (i know i had asked a similer question a couple of days ago but i feel i might get a much more reasonable answer to this one..may be the way i had put accross the question wasn't sensible)

13 comments

[ 2.9 ms ] story [ 45.0 ms ] thread
mfukar
a = a (equality is reflexive under Peano arithmetic)

=> a + c = a + c (addition is commutative under Peano arithmetic)

=> a + c = b + c (a == b)

=> a + c = b + d (c == d)

qed

gdl
Informal reasoning: in a=b, the 'a' and 'b' are merely placeholders that act as pointers to the same underlying value, so they can be used interchangeably. It's no different than referring to "seven" and "the integer directly following six" - I could use either reference any place I could use the other with no effect on the overall statement. Ditto for c=d. All that is happening with a+c=b+d is that two references to values are being changed to different references with the exact same underlying values, so the result is necessarily the same.

See also http://en.wikipedia.org/wiki/Peano_axioms if you prefer the math and logic jargon. High school geometry taught me to dislike dealing with formal proofs, but I think that should be about the right area to look.

pencil
i know even i used to hate formal proofs!!!i assumed this might have a formal proof by not knowing that this property is based on informal reasoning. (but i personally like formal proofs!!!!!)
KoZeN
if a=b,c=d then a+c=b+d

Also, d= a-b+c

Probably doesn't help but thats about the limit of my capabilities!

pencil
we are in the same page !!!!
[dead] startupgrrl
You don't need any properties of the reals, just some well-defined operator (+) and equality (=).

a+c = a+c (reflexive property of equality) a = b => a+c = b+c (substitution axiom) c = d => a+c = b+d (substitution axiom) a+c=b+d. (Done)

TomK32
a = b c = d

a + c = b + d a + c = a + c # replaced b with a and d with c

pencil
oh ya..that looks like a formal proof!!!!!!!!!
sleepdev
Tangentially related question: how are real numbers formally defined?

I remember that integers are usually defined in terms of successors: Succ 1 = 2. But this doesn't help for real numbers because they can't really be enumerated?

pencil
well..this gives the definition of real numbers http://en.wikipedia.org/wiki/Real_numbers..but doesn't mention even a single bit about the truthfullness of the properties of realnumbers.(to be honest i'am not in a position to come out with a rational explanation!!!!!)
vladoh
Ohhhhh NOOO... my brain evaporated because of the stupidity of this thread...
pencil
no this isn't stupid.
patio11
There are a bunch of ways to prove this. Let's start with addition, subtraction, equality being commutative, that a + 0 = a, and that a - a = 0

Suppose that d != a - b + c. Since a = b, a - b = 0. This implies d != c. This is a contradiction, so our supposition is inaccurate. d = a - b + c

Now, suppose a + c != b + d. Plug in what we just learned. a + c != b + a - b + c. The b's cancel, leaving another contradiction. Thus, supposition inaccurate, so a + c = b + d. QED