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chosenmaths3435.bsky.social
Throw away most fractions. Nothing changes.. #math #RationalNumbers #OrderIsomorphism #BackAndForth
November 7, 2026 at 2:57 AM
chosenmaths3435.bsky.social
There are exactly as many fractions as integers. #math #Countable #SetTheory #RationalNumbers
November 7, 2026 at 2:57 AM
mathify-dev.bsky.social
Why Defining Integers Inside Rationals Is So Hard

The equation 2x − 1 = 0 exposes the gap immediately: x = 1/2 is rational, but it is not an integer. So undecidability over integers does not automatically transfer to rational solutions.

#hilbertstenth #rationalnumbers #logic #numbertheory
November 7, 2026 at 2:02 AM
mathify-dev.bsky.social
Why Integer Zeros Fit Through a Finite Corridor

A bounded search over integers really can become a finite checklist. If a problem supplies a bound B, then any relevant integer zero of f(n) = 0 must be among the values from −B to B.

#hilbertstenth #integers #rationalnumbers #computability
November 7, 2026 at 2:01 PM
mathify-dev.bsky.social
Why Rational Solutions Change Hilbert’s Tenth Problem

An equation can have integer roots, rational-but-not-integer roots, or no rational roots at all. For x² − 4 = 0, the solutions are −2 and 2, so integers work.

#hilbertstenth #rationalnumbers #undecidability #numbertheory
November 6, 2026 at 6:02 PM
puzzlevai.bsky.social
0.75 stops. 0.333… repeats.

Both are rational because both equal fractions:

0.75 = 3/4
0.333… = 1/3

A rational decimal either terminates or eventually repeats a fixed pattern. That’s the rule.

#RationalNumbers #Math8 #PuzzleVAI
September 21, 2026 at 4:28 PM
puzzlevai.bsky.social
A rescue boat starts with 18 liters, uses 7, then receives 4 more.

18 − 7 + 4 = 15 liters.

The Grade 7 skill is translating each change correctly: fuel used means subtract; fuel received means add.

#Grade7Math #RationalNumbers #PuzzleVAI
August 26, 2026 at 8:16 PM
puzzlevai.bsky.social
Why does 1 ÷ 3 never end?

⅓ = 0.333…

The same remainder returns during long division, forcing the digit to repeat forever. Every rational number has a decimal that either terminates or eventually repeats.

#Grade7Math #RationalNumbers #PuzzleVAI
August 25, 2026 at 10:41 PM
yrroi.bsky.social
Notes from the sqrt(2) proofs stream from 2025 08 03. Check out the 2 min preview for free, and the original 1hr 20min VOD available for patreon members in the quoted post below. #math #notes #proof #squareroot #root2 #irrational #rationalnumbers
A rational number can be expressed as a fraction or ratio of integers: a over b, for any a and b that are integers. Prove sqrt(2) is irrational (proof from discover maths): Assume the opposite. Fundamental theory of algebra is that every natural number larger than one can be expressed uniquely as a prime factorization. assume sqrt(2) = a/b is irreducible for any a and b larger than zero. The expression above can be squared on both sides, and reduced to 2b^2 = a^2. Because a/b must be irreducible a and b are both prime, meaning we have two different prime factorizations of the same number. Proof by contradiction, therefore sqrt(2) must be irrational. Second proof (from vihart.): Consider a right angle triangle where the legs are the same length. By the Pythagorean Theorem, the hypotenuse (hyp) must be sqrt(2). If the leg is an odd length, the hyp must be odd, but if the hyp is odd then it cannot be split into the two equal leg lengths. If the hyp is even, then two times the leg length squared must also be even, meaning the leg must be even, which means the triangle can be further reduced until the leg length is odd. By this logic the hyp cannot be odd, nor can it be even, therefore it must be irrational. Third proof (from Vsauce ding!): If the number 2x is even, then we can be certain that the number 2x+1 is odd. An even number squared is even. An odd number squared is odd. Assume sqrt(2) = a/b. Squaring that equation gives 2b^2 = a^2. If we assume a is even, then we can write a as 2 times some other number c, and we can write the previous equation as 2c^2 = b^2. From this b^2 must be even, and b must also be even. If a and b are both even, then a/b is not in its simplest form, breaking our original assumption and proving that sqrt(2) cannot be written as any a/b, and is therefore irrational. Fourth proof (from tipping point math): If a and b are any two integers, then a^2 has an even number of 2's as its factors, and 2b^2 has an odd number of 2's as its factors. From this argument, the absolute value of a^2 - 2b^2 must be larger than or equal to one. Absolute value of sqrt(2)- (a/b), multiplied by the absolute value of sqrt(2) + (a/b) over itself (equating to a multiply by 1/1). Multiply both the numerator and denominator by (b^2/b^2). Reduce this expression to the absolute value of (a^2 - 2b^2) over the absolute value of (b(sqrt(2)b + a). Reduce to the absolute value of( one over b(sqrt(2)b + a)), and realize this expression must be larger than 0, proving sqrt(2) cannot be written as a/b for any integers a and b, as originally assumed. Fifth proof (also from tipping point math): Consider a right isosceles triangle where both legs are the same length. if sqrt(2) = x/y then there must be some a and b integers which can describe the hypotenuse a and the leg b in the triangle's lowest form. If a and b are integers, then a-b must also be an integer. Consider a right isosceles triangle embedded within the previous triangle, where the legs of the smaller triangle are both the integer a-b. The hypotenuse of this smaller triangle must be the remaining part of the larger leg length b, subtracted the smaller leg length a-b, simplified to 2b-a. As the smaller triangle is inscribed into the corner of the larger triangle, the ration between leg and hypotenuse must be the same, sqrt(2), and therefore the form a/b cannot be the lowest form of the triangle, therefore sqrt(2) cannot equal any fraction of integers. Sixth proof (from wrath of math, originally demonstrated in 2000 by Tom Apostol of Caltec): Assume sqrt(2) = a/b, which is the same statement as assuming that there exists an isosceles right triangle with integer sides. Consider scaling up the triangle by some factor b. Legs of length 1 will become length b. Hypotenuse of length sqrt(2) will become length a. From our first assumption, we can also assume there exists a smallest version of the isosceles right triangle with integer sides, running form point A to O to B and back to A again, with the right angle at point O. The angle AO must equal the angle BO, which we will call x. Now consider a circle drawn about the point B, with a radius of BO, sweeping from O to intersect with the hypotenuse AB at some point C. Consider a triangle drawn tangential to that circle, with a right angle at point C, drawn from C to A to the intersection of that tangent with the line AO at some point D. This is a smaller isosceles right triangle with integer sides, and we have reached the proof of absurdity.
July 27, 2026 at 2:56 AM
qnfo.mstdn.science.ap.brid.gy
Q-based physics: rational numbers are sufficient for all finite-precision measurements. Four sub-claims, four falsification conditions, direct response to the base-field objection. DOI: 10.5281/zenodo.21638963 #qnfo #rationalnumbers #physicsfoundations
July 30, 2026 at 4:47 PM
sciencehumor-io.bsky.social
The Infinity Gap

#Math #Probability #Realnumbers #Rationalnumbers #Infinity

https://sciencehumor.io/math-memes/the-infinity-gap-sx4l
The Infinity Gap
July 15, 2025 at 4:29 AM
purplemath.com
True story! (Well, actually, they took Hippasus out on a lake, where he had "a boating accident".)

#purplemath #mathhumor #Pythagoras #Pythagoreans #rationalnumbers #irrationalnumber #irrationals #Hippasus #rigthtriangle
A four-panel comic, with two squares in each of two rows. Row 1, left-hand image: Two persons are shown, with one (labeled "Pythagoras") standing behind and to the right of the one (labeled "Hippasus") seated at a table looking at a sheet of paper in front of him. Hippasus says, "Hey bro! Look at what I wrote!" Row 1, right-hand image: We see what Pythagoras is seeing by looking over Hippasus' shoulder. A right triangle is shown, having legs of length 1 and thus having an hypotenuse of length the square root of 2. (Pythagoras led a group that believed, with religious fervor, that all numerical values are rational. The existence of an irrational number like the square root of 2 could cause severe cognitive dissonance.) Row 2, left-hand image: We see a close-up of Pythagoras' face, eyes narrowed, as he examines Hippasus' drawing. Row 2, right-hand image: Pythagoras is shown in a throwing stance, ready to hurl something with all his strength. Hippasus is shown, maybe the size of a house-cat, in Pythagoras' left hand, clearing being what Pythagoras intends to toss away as hard as he can. The caption says, "Yeet Hippasus".
September 25, 2025 at 5:11 PM
purplemath.com
Nah, man; that's just *so* wrong!

#English #many #much #language #rationalnumbers #RealNumbers #MathHumor #purplemath
A four-panel comic. Panel 1: A red-shelled snail is following a blue-shelled snail. Red says, "English is difficult. When do we use 'many' and when do we use 'much'?" Panel 2: Blue replies, "'Many' is for nouns that are countable, like apples. 'Much' is used for nouns that are not countable, like water." Panel 3: Red then asks, "So we should say 'many rational numbers' and 'much real numbers'?" Panel 4: Blue stops, turns around, and stares (glares?) silently at Red.
January 22, 2025 at 4:33 PM