+0

# Is 2 equal to 1 ?

+5
477
7
+869

$$\\a=b\\ a\times a= a\times b\\ a^2=ab\\ a^2+a^2=a^2+ab\\ 2a^2=a^2+ab\\ 2a^2-2ab=a^2+ab-2ab\\ 2a^2-2ab=a^2-ab\\ 2a^2-2ab=1a^2-1ab\\ 2(a^2-ab)=1(a^2-ab)\\ 2=1$$

Can you spot the miscalculation ?

EinsteinJr  Apr 30, 2015

#7
+869
+5

For those who haven't understood, the miscalculation is on the penultimate line :

Since a²=ab, a²-ab=0;

And in the 9th line, I divided each side by a²-ab, so I divided by 0; and as you must now, the division by 0 is impossible.

EinsteinJr  May 1, 2015
#1
0

a=b

axa=axb

a^2=ab

a^2+a^2=a^2+ab

a^4=a^2+ab

2a^2-2ab=a^2-ab

2a^2-2ab=1a^2-1ab

the correct answer is to this part is: 2a^2-2ab=a^2-ab

so the final answer is a=square root of ab

Guest Apr 30, 2015
#2
+92775
0

I do not think that you have explained what i wrong with the logic :/

Melody  Apr 30, 2015
#3
+869
+5

HINT: Since a=b, what is a-b ?

EinsteinJr  Apr 30, 2015
#4
+92775
0

Yes, that is a good hint :)

Melody  Apr 30, 2015
#5
+869
0

Did you spot the miscalculation Melody ?

EinsteinJr  Apr 30, 2015
#6
+92775
0

Yes i did :)

Melody  May 1, 2015
#7
+869
+5

For those who haven't understood, the miscalculation is on the penultimate line :

Since a²=ab, a²-ab=0;

And in the 9th line, I divided each side by a²-ab, so I divided by 0; and as you must now, the division by 0 is impossible.

EinsteinJr  May 1, 2015