View Full Version : Do numbers lie, redux
Arctic Fox
02-21-2013, 07:25 PM
let X+Y=Z
multiply both sides by four:
4X+4Y=4Z
and by five:
5Z=5X+5Y
add the two:
4X+4Y+5Z=5X+5Y+4Z
subtract 9Z from each side:
4X+4Y-4Z=5X+5Y-5Z
thus:
4(X+Y-Z)=5(X+Y-Z)
divide both sides by (X+Y-Z):
4=5
DougB
02-21-2013, 07:58 PM
...
When you get to 4(X+Y-Z)=5(X+Y-Z) you are multiplying 4 and zero which equals zero. 5 multiplied by zero is also zero....so what you have is 0=0 not 4=5
skyguy79
02-21-2013, 09:54 PM
Does a cross-dressing mathematician wear an algebra? :coolsmiley::1rotfl:
Arctic Fox
02-22-2013, 07:50 AM
When you get to 4(X+Y-Z)=5(X+Y-Z) you are multiplying 4 and zero which equals zero. 5 multiplied by zero is also zero....so what you have is 0=0 not 4=5
Top of the class, Pooh :-)
Uskesso
02-26-2013, 05:32 AM
Does a cross-dressing mathematician car parts (http://www.robustbuy.com/car-accessories-c-492.html) wear an algebra? :coolsmiley::1rotfl:
Worst knowledge of algebra. Must not become maths teacher
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