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Today is the birthday of Augustin Fresnel born 1788 in Broglie, France. Fresenel did important work on optics where he was one of the founders of the wave theory of light. At age twelve Fresnel began his studies at the École Centrale in Caen. He was introduced to science and he began to show a liking
for mathematics. He contributed his interest to the inspiration provided by his teachers. He decided on a career in engineering. He entered the École Polytechnique in Paris in 1804 where he began his analysis on optics. He often quoted that "Nature is not embarrassed by difficulties of analysis."
Inspiration, passion, perseverance, and enjoyment are ingredients that are blended in the biographies of many mathematicians.
Another story of a moment that I knew that mathematics was my chosen field. I had just started my masters degree in mathematics. I had put off this journey for some time. I had doubts about my ability in mathematics but I had recently found a paper I had written as an undergrad to gain acceptance into a student teaching position. In this paper, I had listed a number of life goals, one of which was obtaining my masters degree by the age of 40. I was 35 at the time. Also, I was commuting to work with a friend who had started his graduate work and he prodded me, relentlessly so I began the journey.
One of the my early classes was on teaching geometry and in this class the students were required to solve several proofs. A proof that was assigned was:
Given a convex quadrilateral and midpoints on each side, prove that the midpoints are the vertices of a parallelogram.
The following is my proof:
1. Segment FB and segment EC are parallel because of the triangular midline theorem.
2. Segment HG and segment EC are parallel because of the triangular midline theorem.
3. Segments FB and EC are parallel because of the transitive property.
4. The length of FB is 1/2 the length of EC as is HG by the triangular midline theorem.
5. Thus FB and HG are of equal length and parallel.
6. Quadrilateral FHGB is a parallelogram since opposite sides congruent and parallel is a sufficient condition.
I had worked on this proof the next day and finally solved it as I was driving on the way home. I was literally driving and deriving :-) When I completed the proof, I looked up and saw that a milk truck was barreling in my direction in the lane I was occupying - the left lane. I immediately corrected course and avoided the collision, however, I felt at peace. I was ecstatic by my solution and was resigned to any resulting consequence, even one that may have been considered by others as a tragic end. I was the only one the next week that had determined the proof. Obviously, my fellow students disapproved of solving math problems as they commuted.
Inspiration, passion, perseverance, and enjoyment coincided at that moment and resulted in a powerful moment I will remember for a long time.
Another story of a moment that I knew that mathematics was my chosen field. I had just started my masters degree in mathematics. I had put off this journey for some time. I had doubts about my ability in mathematics but I had recently found a paper I had written as an undergrad to gain acceptance into a student teaching position. In this paper, I had listed a number of life goals, one of which was obtaining my masters degree by the age of 40. I was 35 at the time. Also, I was commuting to work with a friend who had started his graduate work and he prodded me, relentlessly so I began the journey.
One of the my early classes was on teaching geometry and in this class the students were required to solve several proofs. A proof that was assigned was:
Given a convex quadrilateral and midpoints on each side, prove that the midpoints are the vertices of a parallelogram.
The following is my proof:
1. Segment FB and segment EC are parallel because of the triangular midline theorem.
2. Segment HG and segment EC are parallel because of the triangular midline theorem.
3. Segments FB and EC are parallel because of the transitive property.
4. The length of FB is 1/2 the length of EC as is HG by the triangular midline theorem.
5. Thus FB and HG are of equal length and parallel.
6. Quadrilateral FHGB is a parallelogram since opposite sides congruent and parallel is a sufficient condition.
I had worked on this proof the next day and finally solved it as I was driving on the way home. I was literally driving and deriving :-) When I completed the proof, I looked up and saw that a milk truck was barreling in my direction in the lane I was occupying - the left lane. I immediately corrected course and avoided the collision, however, I felt at peace. I was ecstatic by my solution and was resigned to any resulting consequence, even one that may have been considered by others as a tragic end. I was the only one the next week that had determined the proof. Obviously, my fellow students disapproved of solving math problems as they commuted.
Inspiration, passion, perseverance, and enjoyment coincided at that moment and resulted in a powerful moment I will remember for a long time.
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