Simple Math Problem DEBATE

Inversely · Jan 23, 2018

Slayed said: ↑

To me at least, I found the whole relativity / space side of physics to be the most interesting by far.
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that was the gayest thing ive heard you say yet

automatik · Jan 23, 2018

Inversely said: ↑

that was the gayest thing ive heard you say yet
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reported

Slayed · Jan 23, 2018

Inversely said: ↑

that was the gayest thing ive heard you say yet
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Says the kid who dropped out of accel phys cause it was too hard

Inversely · Jan 23, 2018

Slayed said: ↑

Says the kid who dropped out of accel phys cause it was too hard
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84>83.5

Slayed · Jan 23, 2018

Inversely said: ↑

84>83.5
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86 >> 40

gun · Jan 23, 2018

Inversely said: ↑

that was the gayest thing ive heard you say yet
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says the one who thinks drugs are cool

Ultraviolet · Jan 23, 2018

These posts are weird solutions the maths problem :/. Let's get back on track and solve it. Here was my answer:

Ultraviolet said: ↑

ItsColinGuys Ok, so 'a', 'b' and 'c' must be integers. I pictured the cross-section of a cuboid and the lengths (as you called 'g') is √(a^2+√((b^2)+(c^2))). As you said, numbers like 3, 4 and 5 fit nicely into the Pythagorous theorem. However, since we are talking about 3-d objects, not only should √(a^2+√((b^2)+(c^2))) equal an integer but √(c^2+√((b^2)+(a^2))) must also equal that as well. I have made a graph: (b^2)=(a^2)+(x^2) and y=(a^2)+(x^2) to found out where they intersect. a is the first length of a triangle and b is the hypotenuse of the triangle. You can see that when b=5 and a=3, x must equal 4. Or when b=13 and a=12, x must equal 5. This is the way to find out perfect triangles where all the sides are integers. However, you cannot keep change the hypotenuse with opposite side or the adjacent side and have the new hypotenuse equal an integer. Like in the example, 3^2+4^2=5^2. You can't expect 5^2+3^2 to equal an integer. Therefore, the last part of √(a^2+√((b^2)+(c^2))) -> √((b^2)+(c^2))) cannot be an integer. So √a^2+... must equal an integer. You can't square an integer, add it to an irrational number, then take the square of that number and expect that number to be an integer.

In conclusion: The closest you can get to solving this problem is to have a, b, and c as integers and have two 'top corner to bottom corners' be integers. You can't get closer than that. Hopefully you can understand all this...
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Inversely · Jan 23, 2018

gun said: ↑

says the one who thinks drugs are cool
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Says the one who’s to pussy to do anything

Examples · Jan 23, 2018

***** wha

Liamhw14 · Feb 16, 2019

Ultraviolet said: ↑

Where in New Zealand are you? I live in Auckland
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I Live I Auckland

octo · Feb 17, 2019

how many prime num ers are their? more or less then 1,0000,00?

edit: just in case u dont know a prime number is a number that cant be devided..

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Simple Math Problem DEBATE

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