Math+Physics major only

Colleges & Universities Massachusetts Institute of Technology
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<p>Let’s prove some irrational numbers to start. =D</p>

<p>I will start.</p>

<p>Assume that sq.rt.2 is rational.
sq.rt.2 can be expressed by p/q as a simplest form where p and q belong to Z.
sq.rt.2 = p / q
2 = p2 / q2
p2 = 2q2
p2 is an even number.
p is an even number.
p can be expressed by 2s where s belong to Z.
q2 = p2 / 2
q2 = 4s2 / 2
q2 = 2s2
q2 is an even number.
q is an even number.
p/q can be further reduced.
p/q is not the simplest form of sq.rt.2.
This contradicts what we have assumed at the beginning of this proof and so the assumption is false.
sq.rt.2 is not a rational number.
sq.rt.2 is an irrational number.</p>

<p>I have found 4 proofs so far. I believe there is more. =D</p>

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<p>… why is this math/physics majors only?</p>

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<p>What exactly are you going for here? Also, why is it in MIT? Wouldn’t you nail this up in some other more cloely fitting catagory.</p>

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<p>Oh so you weren’t there. =D But hey, you have to be a monkey to understand all this. Eh? =D</p>

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<p>Um?</p>

<p>I remember doing this in grade 9. What’s so Math+Physics-requiring in this?</p>

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<p>me too, but why here?</p>

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<p>MrPropaganda - Learn LaTeX if you have not, although you cannot post LaTeX here.</p>

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<p>LaTeX is easy. Use TeXnicCenter and read the Wikibooks.</p>

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<p>@ MrPropapanda
I have irrational related problem for you.</p>

<p>Prove that there are infinitly many irrational numbers?</p>

<p>Hint: You can use similar steps as above.</p>

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<p>Prove by contradiction:
Assume that there is a finite amount of irrational numbers.
There will be a largest irrational number p…</p>

<p>Prove by contrapositive:
Prove that there are an infinite number of rational numbers and that there is at least one irrational number between two rational numbers…</p>

<p>I’m stumped. Please do explain.</p>

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<p>I’m not going to major in Math or Physics, but may I join? :D</p>

<p>1/ From MrPanda’s idea:
“Prove that there are an infinite number of rational numbers and that there is at least one irrational number between two rational numbers…”
This is a great idea. So I’ll start with this one. But I’m not sure whether we count negative numbers in the rational and irrational sets, so I’ll just consider positive numbers.</p>

<p>Since sqrt(2) is proved to be irrational, we may use this for further use.
_ If z is rational, then z’=z.sqrt(2) is irrational. Indeed, if z’ is rational then since z is rational, sqrt(2) = z’/z is rational!
_ Take any two rational numbers a1/a2 and b1/b2, where a1, a2, b1, b2 are natural numbers and a1/a2 <= b1/b2. If we can prove that there is at least a rational number z that a1/a2 <= z.sqrt(2) <= b1/b2, then the problem is solved.
_ Because a1/a2 = (a1.b2)/(a2.b2) <= b1/b2 = (b1.a2)(a2.b2), we must have: a1.b2 <= (b1.a2 - 1).
_ Thus, the condition above becomes: a1.b2 <= (b1.a2 - 1) <= a2.b2.z.sqrt(2) <= b1.a2.
_ The problem now becomes that we have to find a rational number Z that satisfies:
A-1 <= Z.sqrt(2) <= A,
or A/sqrt(2) - 1/sqrt(2) <= Z <= A/sqrt(2),
where A is a natural number.
_ Since 1/sqrt(2) >0.6, denote B=A/sqrt(2), we have: B-0.6 <= Z <= B.
_ If B-<=0.6, then we should find Z where: B-0.6<=<=Z<=B. Of course, <=B, so we can take Z= in this case.
_ If B-**>0.6, since B-1 < ** <= B, we should find Z where: B-0.6 < +C <= Z <= B, where C is a rational number between 0.4 and 0.6. So we can take Z=+C.
The problem is solved.</p>

<p>2/ From Inconclusive’s hint: “You can use similar steps as above.”
I don’t know which way you are guiding us to, but here is my idea, based on your hint: we can use similar method to prove that if z is a rational number then z’=z.sqrt(2) is irrational. However, we’ve already proved that sqrt(2) is irrational, and with this fact, we’ve just proved this in the solution above :smiley: Because there is an infinite number of rational numbers, there is also an infinite number of irrational numbers whose form is similar to z’.</p>

<p>3/ From MrPanda’s idea #1:
“Assume that there is a finite amount of irrational numbers.
There will be a largest irrational number p…”
Because z.sqrt(2) is irrational and the set of rational numbers is infinite, we can always find a rational number z such that z.sqrt(2)>p (shown) :D</p>

<p>P.S.: Why is this thread only for those whose major is Math/Physics???</p>

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<p>“_ Take any two rational numbers a1/a2 and b1/b2, where a1, a2, b1, b2 are natural numbers and a1/a2 <= b1/b2. If we can prove that there is at least a rational number z that a1/a2 <= z.sqrt(2) <= b1/b2, then the problem is solved.
_ Because a1/a2 = (a1.b2)/(a2.b2) <= b1/b2 = (b1.a2)(a2.b2), we must have: a1.b2 <= (b1.a2 - 1).”</p>

<p>Sorry, there are a few errors. It should be:
“_ Take any two rational numbers a1/a2 and b1/b2, where a1, a2, b1, b2 are natural numbers and a1/a2 < b1/b2. If we can prove that there is at least a rational number z that a1/a2 <= z.sqrt(2) <= b1/b2, then the problem is solved.
_ Because a1/a2 = (a1.b2)/(a2.b2) < b1/b2 = (b1.a2)(a2.b2), we must have: a1.b2 <= (b1.a2 - 1).”</p>

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<p>Alright… A potential math major here :)</p>

<p>Now sqrt2 is the root of the equation x^2-2=0</p>

<p>If there is a rational root for this equation, say, p/q, then p divides 2 and q divides 1. By exhausting all the possibilities, we can see that no p/q fulfill this requirement, hence, sqrt2 is irrational.</p>

<p>@12npm12: What’s your intended major? I thought you are inclined to physics…</p>

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<p>@kenhungkk: I actually like physics, not … physics :smiley: To me studying physics is learning the perspectives to approach this world. I like thinking of how Newton could deduce his laws, what force really is, etc, not solving problems (that’s why I suck at doing problems involving lots of calculation :D). And so I haven’t decided on my major yet. It might be Aeronautical Engineering, which is my dream of childhood, or business or something socially-related, since I would like to learn the perspectives :)</p>

<p>P.S.: Your solution is great :smiley:
I intended to start with the Pythagore theorem, but it turned out to be basically the same as the popular solution that MrPanda introduced.</p>

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<p>@12npm12: A MathLinker I am :slight_smile: Indeed, I have seen only MrPropapanda’s proof in the past.</p>

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<p>@Inconclusive:</p>

<p>Take sqrt(2). It’s irrational. Now divide by 2. Another irrational. Now keep doing that - infinitely many irrationals.</p>

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<p>@nikki93: You must prove that sqrt(2)/2^n is irrational.</p>

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<p>Fallacy in your proof nikki93. There are times when a rational number multiplied by an irrational number becomes a rational number.</p>

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<p>@MrPanda: Example?</p>

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<p>@12npm12: 0 times pi = 0.
@MrPropapanda: nikki92’s idea is viable. All the proof needs is a minor edit.</p>