Root 2 - Numberphile
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Date: 2022-04-08
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Comments and reviews: 9
Martin
5: 30 Let me fill in a proof:
Assume the number i doesn't exist!
Set x = -2
then x-2 = 4
in other words
x-2 = 2-2
take the logarithm on both sides
log (x-2) = log (2-2)
Use logarithm rules.
2log x = 2 log 2
Divide by 2
log x = log 2
Cancel the log
x = 2
But we assumed that
x = -2
So this leads to
-2 = 2 -> 0 = 4
In other worlds:
4 = 0
A contradiction!
reply
5: 30 Let me fill in a proof:
Assume the number i doesn't exist!
Set x = -2
then x-2 = 4
in other words
x-2 = 2-2
take the logarithm on both sides
log (x-2) = log (2-2)
Use logarithm rules.
2log x = 2 log 2
Divide by 2
log x = log 2
Cancel the log
x = 2
But we assumed that
x = -2
So this leads to
-2 = 2 -> 0 = 4
In other worlds:
4 = 0
A contradiction!
reply
Kenneth
When we make the proof of (2)-(1/2) with (4)-(1/2) which is 2. Then we get
4 = (a/b)-2
4b-2=a-2
a is dividable by 4 (not by 2 like the number in the Video)
And then lets take a = 4c
so 4b-2 = (4c)-2
4b-2 = 16c-2
b-2 = 4c-2
b is also dividable by 4
so a and b are both dividable by 4 and 2 is irational?
I hope I made my point that I do not understand that way of prove! :(
reply
When we make the proof of (2)-(1/2) with (4)-(1/2) which is 2. Then we get
4 = (a/b)-2
4b-2=a-2
a is dividable by 4 (not by 2 like the number in the Video)
And then lets take a = 4c
so 4b-2 = (4c)-2
4b-2 = 16c-2
b-2 = 4c-2
b is also dividable by 4
so a and b are both dividable by 4 and 2 is irational?
I hope I made my point that I do not understand that way of prove! :(
reply
marcelius
sqrt2=[[[ C ]=[[ c-2 = 1+1 ]=[[ Teorema Academica Martirosianas Marcelius (a+b)-(n+1)=c-n - [[ a+ b= c]
a+b= integral po versijos =O = [a= b] 2b= O b+1=-1 po versijos b=1, trivialiai a=1 ' a=1, b=1 a=o, b=o
ant decartian cordinat kvadratas su tricampis ABC ab=1 bc=1, [ c-2= 1+1 ]= C= sqrt2 = [ 1+1= 2]= [1+1=sqrt2.
reply
sqrt2=[[[ C ]=[[ c-2 = 1+1 ]=[[ Teorema Academica Martirosianas Marcelius (a+b)-(n+1)=c-n - [[ a+ b= c]
a+b= integral po versijos =O = [a= b] 2b= O b+1=-1 po versijos b=1, trivialiai a=1 ' a=1, b=1 a=o, b=o
ant decartian cordinat kvadratas su tricampis ABC ab=1 bc=1, [ c-2= 1+1 ]= C= sqrt2 = [ 1+1= 2]= [1+1=sqrt2.
reply
marcelius
sqrt2=[[[ C ]=[[ c-2 = 1+1 ]=[[ Teorema Academica Martirosianas Marcelius (a+b)-(n+1)=c-n - [[ a+ b= c]
a+b= integral po versijos =O = [a= b] 2b= O b+1=-1 po versijos b=1, trivialiai a=1 ' a=1, b=1 a=o, b=o
ant decartian cordinat kvadratas su tricampis ABC ab=1 bc=1, [ c-2= 1+1 ]= C= sqrt2 = [ 1+1= 2]= [1+1=sqrt2.
reply
sqrt2=[[[ C ]=[[ c-2 = 1+1 ]=[[ Teorema Academica Martirosianas Marcelius (a+b)-(n+1)=c-n - [[ a+ b= c]
a+b= integral po versijos =O = [a= b] 2b= O b+1=-1 po versijos b=1, trivialiai a=1 ' a=1, b=1 a=o, b=o
ant decartian cordinat kvadratas su tricampis ABC ab=1 bc=1, [ c-2= 1+1 ]= C= sqrt2 = [ 1+1= 2]= [1+1=sqrt2.
reply
John
Proof that the square root of two is irrational? How about proving that (n an integer at least 2) the nth root of any integer is either integer or irrational. (It's easier than the standard proof that sqrt(2) is irrational. Just prove that a non-integer rational raised to the nth power is also a non-integer rational)
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Proof that the square root of two is irrational? How about proving that (n an integer at least 2) the nth root of any integer is either integer or irrational. (It's easier than the standard proof that sqrt(2) is irrational. Just prove that a non-integer rational raised to the nth power is also a non-integer rational)
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riccardo
I don't get it. you said at the beginning that you went by contradiction by saying that a fraction existed. but the fact that you want to keep it to minimal terms (no fractions or even numbers) isn't contraddicting anything but your wish for the fraction to be not like that.
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I don't get it. you said at the beginning that you went by contradiction by saying that a fraction existed. but the fact that you want to keep it to minimal terms (no fractions or even numbers) isn't contraddicting anything but your wish for the fraction to be not like that.
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Wayne
Variation of Brady's proof starting w 2a-2 = b-2. The left side has an odd number of 2's in its prime factorization. The right side has an even number. Replace 2 with any prime number and the proof still works. Thus the square root of any prime number is irrational.
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Variation of Brady's proof starting w 2a-2 = b-2. The left side has an odd number of 2's in its prime factorization. The right side has an even number. Replace 2 with any prime number and the proof still works. Thus the square root of any prime number is irrational.
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Michael
-I'm going to asshume you can write it as a fraction-. That is too too funny!
Funnier still perhaps is that I took no notice of that unique pronunciation a couple years ago when I watched this video which we in America say as assume or ah-ssoom
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-I'm going to asshume you can write it as a fraction-. That is too too funny!
Funnier still perhaps is that I took no notice of that unique pronunciation a couple years ago when I watched this video which we in America say as assume or ah-ssoom
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Fahim
I wish somebody helped me to clear my doubt: if sqrt of 2 is irrational, i. e. it cannot be written as a fraction of two integers, how come you manufacture a piece of paper having sides a and b, where the ratio of a and b is sqrt of 2?
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I wish somebody helped me to clear my doubt: if sqrt of 2 is irrational, i. e. it cannot be written as a fraction of two integers, how come you manufacture a piece of paper having sides a and b, where the ratio of a and b is sqrt of 2?
reply
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