0 = 0 + 0 + 0 + 0 + · · ·

= (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) + · · ·

= 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + · · ·

= 1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + · · ·

= 1 + 0 + 0 + 0 + · · ·

= 1

"Legend has it that, around 1703, in letters to contemporary mathematicians, an Italian monk by the name of Guido Grandi (often called Guido Ubaldus) presented this as proof of the existence of God, since it suggested that the universe could have been created out of nothing! What he actually meant isn't clear, but we do know that the brightest minds of the day were unsure how to explain what the problem was. Leibniz at least recognized that the problem was in the ____ line above;"

I left it blank intentionally

This is copied from my Calculus prof notes, he showed is this "proof" in class.

]]>Yes, you are right.

]]>Hi Ashok123

Your steps aren't algebraicly correct. How did you get that a^2-a^2=a(a-a) ?

The only flawed step was when he divided by

. Everything else is correct.]]>Your steps aren't algebraicly correct. How did you get that a^2-a^2=a(a-a) ?

]]>Hey, swim is new here and is wondering if someone could direct him to the classic 1=2 proof? Thanks.

a=a

(a-a)(a+a)=a(a-a)

a+a=a

2a=a

2=1]]>

Here is another:-

But this proves 2 > 3.

3 < 2

You're missing required grouping symbols, ganesh.

One of the lines above can be fixed by typing:

]]>OK, but it's still a case of:

in post #18, I wrote:

In the example that started this thread, for example, you should learn that dividing by zero is not acceptable in algebra.

except substitute 'number calculations' for 'algebra'.

Bob

]]>Welcome to the forum!

That is a little bit of a twist!

]]>10² = 100 (right)

10²-10²=100-100

(10-10)(10+10)=10(10-10)

So,

(10+10)=10

20=10

2=1

It looks to me like S4 -> S5 is the faulty step.

Just because a^2 = b^2 you may not conclude that a = b eg. 9 = 9 but + 3 is not equal to -3

But if you write

s5) x-(x+y)/2 = -(y-(x+y)/2)

then you get x + y = x + y which seems more reasonable.

Bob

footnote: When I'm trying to track down an algebraic error the following sometimes works.

Choose a value for x and another for y. (Best to avoid 0 and 1 here)

If the value of the LHS = RHS then there's a strong chance the steps to that point are OK.

When LHS not = RHS you know a false step has occurred.

]]>s1) -x.y=-x.y

s2) x^2-x(x+y)=y^2-y(x+y)

s3) x^2-2x(x+y)/2+{(x+y)/2}^2 = y^2-2y(x+y)/2+{(x+y)/2}^2

s4) {x-(x+y)/2}^2 = {y-(x+y)/2}^2

s5) x-(x+y)/2 = y-(x+y)/2

s6) x=y]]>

yes, or 0^0.

Studying false proof can help not making them. Some pitfall are :

a) right deduction, but hypotheses aren't (Sometimes proven after), hence we can deduce nothing

b) the proof uses the result to be proven

]]>That way you learn something about what is allowable in a proof and what isn't.

In the example that started this thread, for example, you should learn that dividing by zero is not acceptable in algebra.

]]>we have not 1=2, except if units are e.g. $ and £ in the year (to find, e.g. in the y 1976, or 1978 http://www.miketodd.net/encyc/dollhist.htm)...1£=2$ were at some time of the story of humanity.

but we were allowed to write

2 is equivalent to 0 (modulo 2)

divide by 2 and

1 is equivalent to 0 (modulo 1).

]]>JaneFairfax wrote:

Jane wrote:Why do you want to learn incorrect proofs? Much better if you just concentrate on learning correct proofs. That might help you pass your exams.

She is still right.

She is even more correct!

Welcome to the forum!

]]>