MathIsCool

[math]ln(2)=\int_{1}^{2}{\frac{1}{t}dt}[/math]

You'll love this: Let [math]a=e^{\frac{ln(2)}{2}}=\sum\limits_{n=0}^{\infty}{\frac{(ln(2))^n}{n!2^n}}[/math]. Then [math]a^2=(e^{\frac{ln(2)}{2}})^2=e^{ln(2)}=2[/math].

Hahahahahahahahahahahahahahahahahahaha.

Like I made an error. This is so funny. The number you're seeking is [math]a=\sqrt{2}[/math]. I'm not going to bother explaining this to you because this place is for mathematics, not for people who don't know 7th grade stuff and babble about pseudomathematical nonsense. My advice would be to find out how [math]\mathbb{R}[/math] is defined. Funny taking insults from people who don't know what they're talking about. Do you know Pythagoras' theorem? If you have a square with lengths 1, it follows for the length a of the diagonal that [math]a^2=1^2+1^2=2[/math]. The only hard fact is that you're completely wrong. I don't know why I'm actually debating with you because it's a hard fact that [math]\sqrt{2}[/math] exists. Like you could argue about a fact. What's your mathematical background?

What are you talking about? [math]0+0=0[/math] is obviously true. From this it follows that 0 is the additive inverse to itself. Since -0 is defined to be the additive inverse to 0, and additive inverses are unique in [math](\mathbb{R},+)[/math], it follows that [math]-0=0[/math]. The uniqueness of inverses is sometimes even included in the definition of a group, yet it follows directly from the group axioms.

It does exist, and it's exactly the length of the diagonal in the unit interval. By the way, in [math]\mathbb{R}[/math], there is never such a thing as a "closest number". You obviously don't know what you're talking about. Often people have trouble understanding that they are wrong. The "intuitive" explanation you give is wrong, as has been pointed out. Also, irrational numbers exist, they aren't approximation tools. Please study the definition of [math]\mathbb{R}[/math]. You have yet to reply to my question above, what's wrong with [math]\pi := \int_{0}^{1} \frac{4}{x^2+1} dx[/math]? Now please don't reply with yet another pointless post but try to address the points I made.

Even easier: [math]0+0=0 \implies -0=0[/math] due to the uniqueness of the additive inverse.

I assume you're talking about the cyclic group [math]C_{60}=\{0,1,...,59\}[/math]. Then the generators are exactly the [math]a\in C_{60}[/math] with [math]gcd(a,60)=1[/math]. Thus there are [math]\varphi(60)=\varphi(3)\varphi(4)\varphi(5)=2\cdot2\cdot4=16[/math] generators.

[math](A\implies B):\iff (\neg A\vee B)[/math]. Implication [math]A\implies B[/math] is defined to be true iff A is false or B is true.

I'm wondering whether he actually knows what [math]\mathbb{R}[/math] is... no offence by any means.

You have to use properties of [math]S_7[/math]. In particular, every [math]s \in S_7[/math] can be written as the product of disjoint cycles, and then the order of s is the lcm of the cycle lengths. Example: [math]s=(1 2 3)(5 6)[/math]. Since the cycles are disjoint, we have [math]o(s)=lcm(3,2)=6[/math]. Now consider any element [math]s \in S_7[/math]. If we wanted the order of s to be 20 (for example), we'd only be allowed to use cycle lenghts 1, 2, 4, 5, 10, 20. Since 10 and 20 are too large (we're in [math]S_7[/math] after all), that means we can only use cycle lengths 2, 4 and 5 (not counting 1 as a cycle length). But out of those, we'd have to use both 4 and 5 (to obtain a lcm of 20) - but we can't, since the cycles have to be disjoint and [math]4+5>7[/math]. So there is no element of order 20 in [math]S_7[/math]. You do the rest.

[math]-0=0+(-0)=0[/math] First identity holds for 0 is the additive neutral of the real field, second holds because a+(-a)=0 for every a.

connector, please don't even try to deny that [math]\sqrt{2}[/math] is a real number. Do you know how the reals are constructed/defined? By the way, if it makes you happy: [math]\pi=\int_{0}^{1} {\frac{4}{x^2 +1} dx}[/math]. Take that as a "definition" and explain where your problem is.