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September 10, 2017 at 10:38 pm (This post was last modified: September 10, 2017 at 10:46 pm by Kernel Sohcahtoa.)
Hello everyone. I was working on a proof yesterday and thought that it was cool enough to share here. Hence, the proposition and my direct proof of it is below.
With that said, in creating this thread, it was my intent for people to be able to post whatever beauty they encountered in mathematics for as long as possible. Thus, I apologize if my post violates community guidelines or causes any inconvenience to the staff and/or AF members.
Proposition: For every four real numbers a,b,c, and d, sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd.
Note the proposition can be re-written in “if-then” form, which yields the following equivalent statement: “if a,b,c, and d are real numbers, then sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd.
Proof.
Let a,b,c, and d be real numbers. Then ad-bc is also a real number. Now, the square of any real number is non-negative, so (ad-bc)^2 ≥ 0 (1). Expanding the left-side of inequality (1) gives (ad)^2 - 2abcd + (bc)^2 ≥ 0 (2). Adding the real numbers (bd)^2, (ac)^2, and 2abcd to both sides of inequality (2) gives (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 ≥ (ac)^2 + 2abcd +(bd)^2, and rewriting the left-hand side of the foregoing inequality via exponential properties yields, a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 ≥ (ac)^2 + 2abcd +(bd)^2 (3). Factoring both sides of inequality (3) gives (a^2 + b^2)*(c^2 + d^2) ≥ (ac + bd)^2 (4). Taking the square root of both sides of inequality (4) yields sqrt((a^2 + b^2)*(c^2 + d^2)) ≥ sqrt((ac + bd)^2), or equivalently (via properties of radicals) , sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd. Hence, the proof is complete.
Notes
[1] The analysis of proof for this proposition was pretty cool. Essentially, I worked backwards from the conclusion that sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd, and via algebra, I ended up reaching the statement that (ad-bc)^2 ≥0 , which was my last statement in the backward process. As a result, I worked forward from the hypothesis that a,b,c, and d are real numbers. I then generated the following forward statements: (1) ad-bc is a real number; (2) the square of any real number is non-negative, so (ad-bc)^2 ≥0. Thus, the last statement in the forward process is precisely the last statement in the backward process, which indicated to me that I had a proof.
[2]This proposition can also be proven via the contradiction method. To proceed with contradiction, we would take the negation of the conclusion that sqrt(a^2 + b^2)*sqrt(c^2 + d^2) ≥ ac + bd, which is sqrt(a^2 + b^2)*sqrt(c^2 + d^2) < ac + bd. We then work forward from this new statement and the hypothesis in order to generate a contradiction. Also, if you work backward from my steps in the direct proof (taking into account the different inequality sign), then via algebra, you should be able to reach the following contradiction:
This morning, I completed a proof of the following proposition: there exist four distinct real numbers a,b,c,d such that exactly four of the numbers ab,ac,ad,bc,bd,cd are irrational. (IMO, this proposition is pretty cool). Note, the proof method to use here is proof by contradiction. Well, happy proof-writing AF members and anyone else.
Hint 1
In order to use contradiction, we must first take the negation of the conclusion that, there exist four distinct real numbers a,b,c,d such that exactly four of the numbers ab,ac,ad,bc,bd,cd are irrational (or equivalently, not rational). Thus, we can negate the statement via the following steps (we distribute the negation symbol through the statement from left to right by first tackling the quantifier "there is" and then tackling "the something that happens" in the statement. Note, the properties themselves stay the same):
i) ~[There exist four distinct real numbers a,b,c,d such that exactly four of the numbers ab,ac,ad,bc,bd,cd are not rational]
ii) For every four distinct real numbers a,b,c,d, ~[exactly four of the numbers ab,ac,ad,bc,bd,cd are not rational]
iii) For every four distinct real numbers a,b,c,d, exactly four of the numbers ab,ac,ad,bc,bd,cd are rational (taking the negation of a statement with the word "not" in it cancels out the "not"). Now, for simplicity, lets refer to this statement as A1, which is a new statement in the forward process: this process consists of statements that we create via facts and mathematical reasoning, which we can work forward from in order to establish the truth of the conclusion (note reasoning backwards from the conclusion via key questions, definitions, facts, etc., is referred to as the backward process.). Thus, since our proof is by contradiction, we must work forward from A1 in order to reach a contradiction (we do not have the luxury of working backward from anything).
Hint 2 (Analysis of Proof/Spolier)
A1: For every four distinct real numbers a,b,c,d, exactly four of the numbers ab,ac,ad,bc,bd,cd are rational.
Now, since, the keywords "for every" appear in A1, we can select any four distinct real numbers, say a=a',b=b',c=c',and d=d'. Thus, we have that
A2: a', b',c', and d' are four distinct real numbers.
Now, from A1, it follows that
A3: Exactly four of the numbers a'b', a'c', a'd', b'c', b'd', c'd' are rational.
Now, without loss of generality, assume that
A4: a'b', a'd', b'c', c'd' are rational.
Since a',b',c', and d' are distinct real numbers (via A2), then from A4, it follows that
A5: a',b',c', and d' are rational.
Now, the product of any two rational numbers is also rational, so it follows from A5 that
A6: a'b', a'c', a'd', b'c', b'd', c'd' must be rational.
But, A6 contradicts A3, which states that "exactly four of the numbers a'b', a'c', a'd', b'c', b'd', c'd' are rational."
Thus, we have reached a contradiction, which completes the proof (please note that the final product must be written in paragraph form in order to complete the proof-writing process).
P.S. I'm not a math guru/nerd/expert, so I apologize if my writing/reasoning is inelegant.
You have shown it to me before, and it was in relation to math proofs. The site looks pretty cool: I've certainly got a lot more to learn and could always improve. Out of curiosity, is the site being recommended due to poor/incorrect mathematical reasoning/proof-writing in my previous post? Thanks.
P.S. In general, to all AF members, if I've made an error in a proof/mathematical work that I've posted here, then if it's okay with you, could you please let me know, so that I can post a corrected version? Thanks, and I appreciate everyone's participation in this thread. Live long and prosper.
You have shown it to me before, and it was in relation to math proofs. The site looks pretty cool: I've certainly got a lot more to learn and could always improve. Out of curiosity, is the site being recommended due to poor/incorrect mathematical reasoning/proof-writing in my previous post? Thanks.
Actually, I posted this link first and read your proof after.
It's always good to have a bunch of people interested in the same thing discussing these things you're so passionate about... and I think you'll find a much more eager audience there... and, mainly, a lot of teachers.
Personally, I'm not a fan of showing this kind of proposition... my physics background sees no use in them!
From everything you post on AF and what you tell us from your life, you seem like a great person... very sure of himself always carefully pondering your every move.
But this question of yours makes me think that you're a bit unsure of yourself... am I reading incorrectly?
(September 21, 2017 at 10:59 am)Kernel Sohcahtoa Wrote: P.S. In general, to all AF members, if I've made an error in a proof/mathematical work that I've posted here, then if it's okay with you, could you please let me know, so that I can post a corrected version? Thanks, and I appreciate everyone's participation in this thread. Live long and prosper.
(September 21, 2017 at 10:59 am)Kernel Sohcahtoa Wrote: You have shown it to me before, and it was in relation to math proofs. The site looks pretty cool: I've certainly got a lot more to learn and could always improve. Out of curiosity, is the site being recommended due to poor/incorrect mathematical reasoning/proof-writing in my previous post? Thanks.
Actually, I posted this link first and read your proof after.
It's always good to have a bunch of people interested in the same thing discussing these things you're so passionate about... and I think you'll find a much more eager audience there... and, mainly, a lot of teachers.
Personally, I'm not a fan of showing this kind of proposition... my physics background sees no use in them!
From everything you post on AF and what you tell us from your life, you seem like a great person... very sure of himself always carefully pondering your every move.
But this question of yours makes me think that you're a bit unsure of yourself... am I reading incorrectly?
(September 21, 2017 at 10:59 am)Kernel Sohcahtoa Wrote: P.S. In general, to all AF members, if I've made an error in a proof/mathematical work that I've posted here, then if it's okay with you, could you please let me know, so that I can post a corrected version? Thanks, and I appreciate everyone's participation in this thread. Live long and prosper.
Will do, if I ever spot it.
I appreciate your thoughtful response, pocaracas. I didn't realize that there were a lot of teachers on that site along with so many people interested in mathematics: it seems like a literal goldmine of wonderful ideas and meaningful exchanges based in curiosity and friendship.
With that said, as to your question contained within the hide box, I would say that always being open to the possibility of how I may be mistaken, is a trait that I value and try to exhibit; it can certainly spare a person some embarrassing moments, especially if that individual has committed an error.
In your post, you said that you have a physics background. Out of curiosity, are you more interested in applied mathematics or pure mathematics (or do you like both)? Thanks again for your response.
(September 21, 2017 at 1:52 pm)Kernel Sohcahtoa Wrote: In your post, you said that you have a physics background. Out of curiosity, are you more interested in applied mathematics or pure mathematics (or do you like both)? Thanks again for your response.
I'd say I like both, but I prefer the applied variety.
Give me 3 dimensions and I can still visualize the thing.... go higher and things become tricky.
When you analyse an algebraic division Ring, a set E with with 2 operations called '+' and '*' with the axioms of the algebraic goup (E.+) previously proven with commutivity a+b = b+a, the exitence of one element we called 0 where for all x in E: x+0 = x, with the set not being empty or singular;
Given the properties of the operation '*':
1: there exists a member we call u where for all set members a, a*u = a
2: for all members of E, a*b = b*a
3: for all a,b,c in E, a*(b+c) = a*b+a*c
With these in mind prove that a*0=0 and while you're at it prove that 0 is not equal this 'u' mentioned on the axioms.
