+0  
 
+5
1269
7
avatar+870 

Read the following calculations:

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

Can you spot the miscalculation ?

 Apr 30, 2015

Best Answer 

 #7
avatar+870 
+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.

 May 1, 2015
 #1
avatar
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

 Apr 30, 2015
 #2
avatar+118587 
0

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

 Apr 30, 2015
 #3
avatar+870 
+5

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

 Apr 30, 2015
 #4
avatar+118587 
0

Yes, that is a good hint :)

 Apr 30, 2015
 #5
avatar+870 
0

Did you spot the miscalculation Melody ?

 Apr 30, 2015
 #6
avatar+118587 
0

Yes i did :)

 May 1, 2015
 #7
avatar+870 
+5
Best Answer

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

4 Online Users

avatar