math

Novus 4510 Mathematician

A friend gave me this National Semiconductor Novus 4510 Mathematician scientific calculator made in 1976, and I couldn’t resist taking it apart.

It is an interesting calculator, simply by being a Reverse Polish notation (RPN) machine.

On the inside, the hardware it’s pretty simple, with only two IC’s and a 9-digit 7-segment bubble LED display, mounted on a single side fiber glass PCB. The keyboard seems pretty robust and it’s fixed to the front part of the case, while the PCB it’s mounted on the back part of the case, connected through a flat flex cable.

IC1: MM5760N – 24 pin DIP, Slide Rule Processor
IC2: DS8864N – 22 pin DIP, Segment-Digit Parallel-Input Display Driver

novus-4510-pi

novus-4510-inside

novus-4510-back

How 1 equals -1

A funny thing I saw the other day…

(1 – 0)2 = (0 – 1)2  //which is true, then adds square roots on both sides

(1 – 0)2 = √(0 – 1)2  //which again is true… extracting the square roots

1 – 0 = 0 – 1  //forgetting the modulus

1 = –1

Math trick to impress your friends

This is cool to impress someone with your “paranormal” skills.

Put someone to choose a random number, preferably a large one ( 3 or 4 figures ), and he (or she) have to keep this number secret from you and do the following math with it:

xyz = he’s or she’s chosen number ( ex. 315 )

Step1: xyz * 3 ( multiply by 3, ex. 315 * 3 = 945 )

Step2: then take the result and add all it’s figures ( abcd => a+b+c+d , ex. 945 => 9 + 4 + 5 = 18 )

Step3: Repeat Step1 with the new number ( the sum of figures, ex. 18 * 3 = 54 )

Step4: Repeat Step2 ( ex. 5 + 4 = 9 )

Repeat steps 1 and 2 until you get a one figure number.

When the person or persons finished the above algorithm, you tell them the number they’ve got ( which with the algorithm above will be always 9 !), and the question that will appear will sure be “How did you know ?”, … :) try it!

Ancient babylonian math problem

Last week i was watching a documentary that  i wanted to see for some time. it is about this documentary from BBC, story of maths . If you didn’t see it, i recommend it to you.

The documentary has 4 parts, and i believe that in the first part, “The Language of the Universe”, i saw an interesting problem from the time of babylonians.

Like the most math problems of the time is about areas; the modern math has it’s roots to the desire of man to be able to calculate and store information about distances, surfaces and patterns.

Archaeologists found pieces of clay with inscriptions that are believed to be children homework, one of them captivated my attention, it was something like this, ( please excuse my lack of talent for graphic art ) :

Let say we have a square with four circles inside it ( equal radius circles ), all circles are tangent to the square and to one each other. The square has a given length of it sides, let say x, for algebraic purpose.

The question is, what is the area of one circle ?


You will say, “what is so hard ?

We know the length of the square’s side ( x = any pozitive real number ), so we can extrapolate the circle radius ( r = one circle radius), and then calculate  the surface of the circle.” … simple, huh ?!

r = x / 4

Scirc = πr2 = π(x / 4)2 = πx2 / 16 , for a discrete value, let’s say x = 5,  Scirc = π52 / 16 = 0.5625 * π

The fun part is that the ancient babylonians did not have a precise method to calculate the surface of a circle, this was done by considering the circle a polygon with “many” sides.

Other method is to calculate the area of the square, and then subtract the area of the shapes between the circles and the square’s sides ( approximating those to known geometric figures, 8 triangles and a square ), and divide the result by four … quite laborious and not so precise.

Math sophism on Pythagorean theorem

A good math sophism is like an old wine; and if you are like me, then, you would enjoy a “sophism” from time to time. Maths less obvious mistakes are like miss prints on stamps.

An elegant example of such mathematical fallacy is good to keep an idiot busy for couple of days. The one I’ll describe next is just what you need.

It starts from the cos2x + sin2x = 1 which is the trigonometric form of the Pythagorean theorem, and it goes like this:

cos2x + sin2x = 1
cos2x = 1 – sin2x
cos x = (1 – sin2x)1/2
1 + cos x = 1 + (1 – sin2x)1/2
(1 + cos x)2 = (1 + (1 – sin2x)1/2)2
for x = π (pi = 3,14159265…), we have,
(1 – 1)2 = (1 + (1 – 0)1/2)2
0 = (1 + 1)2
0 = 4 //interesting ?!

The mistake is a little hard to notice, but the result should tell you that something is wrong.