Welcome to H3 Maths

Blog Support for Growing Mathematicians

Where was the ancient capital of Mathematics?


Modern theoretical mathematics is a complex and abstract field. It frustrates and annoys secondary school students in math classes, but also provides the basis for all the technological wonders we enjoy today.

Without the incredible mind of a 8th century Muslim mathematician, al-Khawarizmi, the world of math today would look vastly different. But where did al-Khawarizmi live and help develop the basis for modern Algebra that we (sometimes) enjoy today?

Al-Khawarizmi, like many of his colleagues, got to work translating ancient Greek and Indian texts. The knowledge of giants such as Pythagoras, Euclid, and Brahmagupta was the pedestal this new generation of scholars would stand on. But al-Khawarizmi’s contributions only begin with translation of Greek and Hindu texts.  From the great Indian book on math, The Opening of the Universe, al-Khawarizmi adopts the idea of the zero as a number. This opened up a whole new world of mathematical possibilities and complexities.

Using the old Roman numeral system made advanced math next to impossible. With a number system that goes from 0 to 9, al-Khawarizmi is able to develop fields such as algebra, which he initially used to calculate Muslim inheritance laws. He builds more on the geometry of the Greeks, and develops the basic ideas many high school math students can recognize today. Read more and discover the location of this famous Islamic city here.

by posted under Uncategorized | Comments Off on Where was the ancient capital of Mathematics?    

Comments are closed.

Post Support

The graph on the left (Coronavirus) is for a time period of 30 days, while the one on the right (SARS) is for 8 months! Very poor graphical comparison and hardly relevant, unless it is attempting to downplay the seriousness of the coronavirus?

10 x 9 x 8 + (7 + 6) x 5 x 4 x (3 + 2) x 1 = 2020

NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]

H3 Viewers

Skip to toolbar