Sometimes there are many ideas in math that make you go

This note is going to be all about the most obvious facts about math. Post the ones that you think are extremely obvious and vote up your favorites.

I am adding a few to get things moving.

Have fun and be totally obvious!

For those of you who want a link to the image - https://i.imgur.com/Fvnb0Xl.png.

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## Comments

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TopNewestJordan Curve TheoremEvery closed non-self intersecting continuous loop on the plane divides it into an interior and exterior region, i.e. inside and outside the loop. Proofs of it are some of the longest, most difficult ones in mathematics, usually involving algebraic topology.

Axiom of ChoiceGiven any number of non-empty sets, it's possible to pick exactly one member from each of those sets, which makes for a new non-empty set. That is, given a lot of boxes, all of them with things in it, I can get another box, and pick one item from all the other boxes and put them into this box. No proof exists for this one, so it has to be made into an axiom.

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I wish I could upvote you again and again! This is gold!

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$\displaystyle 0\times \pi\times \tau\times \phi\times e = 0$

$\displaystyle\text{MIND = BLOWN}$

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The Pigeonhole PrincipleIf you have things stuffed in containers and if there are more things than containers, then at least one container has more than one thing in it.

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PHP wins in my opinion :D

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agreed

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Reflexive property of equality$a=a$

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The obvious one is

$1 + 1 = 2.$

ON page 357 of Principia Mathematica, they finally conclude

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And then came along Kurt Godel.

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$\displaystyle 1+1 = 2$

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haha! this is what even i thought immediately after reading this post.. :P :P

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Lol, same here!

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Intermediate Value TheoremIf a continuous function $f$ with an interval $[a, b]$ as its domain takes values $f(a)$ and $f(b)$ at each end of the interval, then it also takes any value between $f(a)$ and $f(b)$ at some point within the interval.

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Well, you often get to experience your teacher being Captain Obvious in your school. :P Here's something my "maths" teacher wrote on the blackboard while "teaching" us quadratics:

'If the product of two numbers is zero, one of them must be zero.'

And I was like: you don't say?

What's strange is that we always have to write this exact same line (its Bengali translation) whenever we're solving a quadratic otherwise we get a zero out of five, apparently because guys who think this is too obvious don't have any understanding of maths and just memorize stuff.

Oh, and here's another one. I found this in my eleventh grade "maths" textbook.

Law of trichotomy:Given any two real numbers $a,b,$ either $a>b$ or $a=b$ or $b>a.$If you're thinking this isn't something important, well... I found out that there was an exercise at the end of the textbook which asked to state the law of trichotomy using examples. In fact that question appeared more than once in the eleventh grade final "maths" examination before.

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সমাধান কর

$x^2-5x+6=0$

$\Rightarrow (x-2)(x-3)=0$

দুটি সংখ্যার গুণফল শূন্য হলে এদের মধ্যে কমপক্ষে একটির মান শূন্য হবে।

অর্থাৎ,

হয়, $x-2=0$ অথবা, $x-3=0$.

সুতরাং, $x=2$ অথবা $x=3$.

নির্ণেয় সমাধান $x=2, 3$.

Alt text

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"তোমার স্টেপ জাম্প হয়েছে। $x^2 - 5x + 6 = 0 \implies x^2 - 2x - 3x + 6 = 0 \\ \implies x( x-2) - 3 (x-2) = 0 \implies (x-2)(x-3)=0$ এই স্টেপটা না দেখালে নম্বর কাটা যাবে। " -- my maths teacher

I don't want to live on this planet anymore

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Amio bangali tai tomar betha bujhte perechhi

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For anyone else is wondering what @Sreejato Bhattacharya , @Eddie The Head , and I are talking about, here is a short summary:

Depressing stuff, more depressing stuff, suicidal thoughts and finally more depressing stuff.

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Haha. In a similar vein, when they teach induction in school, they are very pedantic about how the format of the proof should be written up, ranging from "state the proposition, prove the base case, prove the induction step, state the conclusion".

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Yes indeed, that's very annoying. Also, whenever we're given to differentiate something, say $f(x) = x^2 + x + 1$ at $x=3,$ here's what we have to write (translated into English):

$\begin{array}{lll} f'(x) & = \dfrac{d}{dx} (x^2 + x + 1) & \quad \text{substituting } f(x) \\ & = \dfrac{d}{dx} x^2 + \dfrac{d}{dx} x + \dfrac{d}{dx} 1 & \quad \text{(the derivative of the sum of two functions is equal to the sum of the derivative of those functions)} \\ & = 2 \cdot x^{2-1} + 1 \cdot x^{1-1} + 0 & \quad (\dfrac{d}{dx} x^n = n x^{n-1} + \text{ and the derivative of the constant function is zero)} \\ & = 2x + 1 & \quad \text{(simplifying)} \\ f'(3) & = 2 \cdot 3 + 1 & \quad \text{(plugging } x=3 \text{)} \\ & = 7 \quad & \text{evaluating} \end{array}$

If we miss any of these "explanations", we lose marks. In general, the school curriculum is obsessed with jargon and nomenclature instead of mathematics itself.

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@Krishna Ar D'you remember the ultra-sucking method they taught to prove that the sum of all the interior angles of a triangle is 180 degrees! And you'd lose half the marks if you don't mention that "since EF is a line parallel to the base BC of the triangle, and Ab is a transversal cutting the two parallel lines, the interior opposite angles DAB and ABE will be equal."

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Simplifying? Evaluating? Seroiusly?

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There cannot exist an infinitely decreasing sequence of positive integers......

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Prove it!

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Commutative Property of Addition :

If $a,b$ be two numbers, then $a+b=b+a$.

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In $2\text{-dimensional}$ space, given a line and a point not on the line, there is exactly one line passing through that point that is parallel to the given line.

I like this fact because it is so simple and yet the source of so much controversy. See here

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If $a = b$ and $b = c$ , then $a = c$

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Given any two numbers $a,b$, either $a<b$ or $b\leq a$

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All positive odd integers can be expressed as the sum of two non-negative, consecutive integers that are less than or equal to it.

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What about $1$?

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What Finn was trying to say was this:

$2k+1=k+(k+1)$

But he didn't state properly.

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$2k+1 > k+1$ only when $k>0$. He didn't deal with the $k=0$ case properly.

Yes. What he missed isLog in to reply

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smallernon-negative consecutive integers, which isn't true for $1$.Log in to reply

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If $a=b$, then for any function $f$, we have that $f(a)=f(b).$ This same property is the one that we exploit when solving equations. (like when we add the same thing to both sides, if we multiply by the same number...)

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Given any two real numbers $a,b$, there exists a real number $n$ such that $a=b+n$

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Lots of geometric properties are obviously obvious for me. For example...

You don't say?!

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A Line Is A Dot That Goes For A Walk!!!!!!!!!!!!

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Symmetric Property of EqualityIf $a=b$, then $b=a$.

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A tangent is always perpendicular to a circle's radius.

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an odd no. added to an odd no. , gives an even number.............(1)

square of an odd no. is an odd no. ..............(2)

square of an even no. is an even no. ..............(3)

yet, there is no pythogorean triplet, in which : (odd no.)square +(another odd no.) square=(even no.) square!!!!!!!!!!!!!

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An odd square must be congruent to 1 mod 4, and an even square must be congruent to 0 mod 4. A pythagorean triplet with odd + odd = even cannot exist because 1mod4+1mod4=2mod4=/=square

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@Mursalin Habib Congratulations for completing a marvelous 200-day streak!!!!!!

Is this question an adaptation from Quora? Actually, I saw the same question, there??? There are some more questions which are good at Quora, you should add them as well.

The most obvious thing, from my perspective, are the Theorems of Euclid's Geometry and his lemma, I suppose.

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Yes.

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Factorial of 0 is same as factorial of 1. 0 as exponent is overriding (no pun there) everything else, always 1.

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Always? What about $0^0$?

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All chords are perpendicular to a circle's radius and diameter

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Any number times one is itself

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A circle's diameter is a straight line that starts from a point on the circle's edge, goes through the circle's center, and connects to another point on the edge

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A circle has one side and a 360 degrees angle

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every number added to another number gives a number greater than both of them, given that both numbers are positive and niether one of them is zero

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Infinity +infinity =infinity

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One divided by zero not exist....

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$666 = 2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 17^2$

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The beast is coming.

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0 divided by 0 is undefined. The square root of -1 is also undefined. (Duh)

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the square root of -1 is the imaginary unit, not undefined.

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The chord of a circle meets the circle at only one point.

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That's not even true...

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'Tangent' is the word he was looking for I believe.

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4

8-8/765-89412354889546/789-59+551654+46566+5*0=0 lolLog in to reply

Lmfao.. i did not know a lot of these.. thank you for this note so i can start studying them hehehe

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$9^3 + 10^3 = 12^3 + 1^3$

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How is this supposed to be obvious?

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Lol

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Well, it's a really complicated branch of mathematics, taxicab numbers that is, but this is so obvious, as long as you have a calculator or a good memory. :/

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Two knot diagrams of the same knot can always be reached through a finite set of Reidemeister moves.

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Someone went to PuMaC, I see.

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The Kepler conjecture states that no arrangement of equally sized spheres filling space has a greater average density than that of the cubic close packing (face-centered cubic) and hexagonal close packing arrangements. In 1998 Thomas Hales, following an approach suggested by Fejes Tóth (1953), announced that he had a proof of the Kepler conjecture. Hales' proof is a proof by exhaustion involving the checking of many individual cases using complex computer calculations. Referees have said that they are "99% certain" of the correctness of Hales' proof, so the Kepler conjecture is now very close to being accepted as a theorem. Source: Wikipedia.

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$\Huge{\color{#D61F06}{5\neq 5^0 \neq 5^{5^{5^{5^{...}}}}}}$

$\Huge{\color{#20A900}{5^{5^{5^{5^{5^{5^{... \infty}}}}}}}} \color{#3D99F6}{= 6^{6^{6^{6^{6^{6^{6^{...\infty}}}}}}}}$

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