The French mathematician Pierre de Fermat died on the 12th of January 1665.He was born in south of France and became a lawyer after studying law at the university of Toulouse.Fermat was the councillor at the high court of Judicature in Toulouse and mathematics only his hobby.To maintain the status of an amateur and to tease his fellow professional mathematicians ,he provided very little or no proof for his theorems.This naturally led him into fiery debates with his contemporaries ,Descartes and Wallis.Fermat contributed to many branches of Mathematics.In optics he calculated the angle a straw or a spoon appears to be bent at when put in water.
Have you ever observed this ?Try it out!
His most famous theorem called Fermat’s Last Theorem .Fermat stated that you cannot find integers a,b,c (Which are non zero)that satisfy:
a^3 + b^3 = c^3, a^4 + b^4 = c^4,a^5 + b^5 =c^5….a^n + b^n =c^n
for any positive integer n>=3.
For example if a=1 and b=3,then 1^3 +3^3 = 1+27=28=c^3.
But there is no integer c which when cubed ,gives 28.On the page where he wrote down the theorem,he wrote:
“I have a truly marvellousproof of this proposition which this margin is too narrow to contain.”
The proof was only discovered 329 years after fermat’s death,in 1994 ,by Andrew Wiles.
It turned out that fermat’s last Theorem is true for n>=3.
Now we all can accept mathematics is a very interesting and intriguing subject.Many mathematicians world wide have made a lot of discoveries,proved a huge number of theorems ,and have deduced many amzing results.Mathematicians also wanted to have some fun.Therefore they wrote proofs for 2=1.Obviously they had some errors and these are called fallacies.For instance 2=1 is proved in 2 different algebric ways as shown below.
Can you identify which of these steps has the fallacy?
Proof 1:
Let a=b
=>a^2=ab [Multiplying by a on both sides]
=>a^2-b^2=ab-b^2 [subtracting b^2 from both sides]
=>(a-b)(a+b)=(a-b)b [Factorizing both sides ]
=>a+b=b [canceling (a-b) on both sides]
=>b+b=b [substituting a=b]
=>2b=b [simplifying the LHS ]
=>2=1 [Cancelling b on both sides]
Proof 2:
Let a=b
=>a^2=ab [Multiplying by a on both sides]
=>a^2 +a^2=a^2+ab [Adding a^2 on both sides]
=>2a^2=a^2+ab [simplifying the LHS ]
=>2a^2-2ab=a^2+ab-2ab [subtracting 2ab from both sides]
=>2a^2-2ab=a^2-ab [simplifying the RHS ]
=>2(a^2-ab)=a^2-ab [taking out the common factor on LHS]
=>2=1 [Cancelling a^2-ab on both sides]
Have you ever observed this ?Try it out!
His most famous theorem called Fermat’s Last Theorem .Fermat stated that you cannot find integers a,b,c (Which are non zero)that satisfy:
a^3 + b^3 = c^3, a^4 + b^4 = c^4,a^5 + b^5 =c^5….a^n + b^n =c^n
for any positive integer n>=3.
For example if a=1 and b=3,then 1^3 +3^3 = 1+27=28=c^3.
But there is no integer c which when cubed ,gives 28.On the page where he wrote down the theorem,he wrote:
“I have a truly marvellousproof of this proposition which this margin is too narrow to contain.”
The proof was only discovered 329 years after fermat’s death,in 1994 ,by Andrew Wiles.
It turned out that fermat’s last Theorem is true for n>=3.
Now we all can accept mathematics is a very interesting and intriguing subject.Many mathematicians world wide have made a lot of discoveries,proved a huge number of theorems ,and have deduced many amzing results.Mathematicians also wanted to have some fun.Therefore they wrote proofs for 2=1.Obviously they had some errors and these are called fallacies.For instance 2=1 is proved in 2 different algebric ways as shown below.
Can you identify which of these steps has the fallacy?
Proof 1:
Let a=b
=>a^2=ab [Multiplying by a on both sides]
=>a^2-b^2=ab-b^2 [subtracting b^2 from both sides]
=>(a-b)(a+b)=(a-b)b [Factorizing both sides ]
=>a+b=b [canceling (a-b) on both sides]
=>b+b=b [substituting a=b]
=>2b=b [simplifying the LHS ]
=>2=1 [Cancelling b on both sides]
Proof 2:
Let a=b
=>a^2=ab [Multiplying by a on both sides]
=>a^2 +a^2=a^2+ab [Adding a^2 on both sides]
=>2a^2=a^2+ab [simplifying the LHS ]
=>2a^2-2ab=a^2+ab-2ab [subtracting 2ab from both sides]
=>2a^2-2ab=a^2-ab [simplifying the RHS ]
=>2(a^2-ab)=a^2-ab [taking out the common factor on LHS]
=>2=1 [Cancelling a^2-ab on both sides]
