How can 1 = 0?

Yes, the only way involves dividing by zero. If you're interested, here's how to do it.

Take two numbers a and b and put them equal.

thus a=b

Now multifly each side of the equation by a

a^2=ab

Subtract b^2 from each side

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

Now factorise

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

Divide each side by (a-b) {NB (a-b) =0 since a=b so we can't do this!)

a+b = b

Noting that a=b we have

2a=a

which means

2=1

subtracting 1 from each side

1=0

QED (except it isn't because of the illegal step of dividing by zero).

This could involve an operation. But to say 1 = 0 is false. 1 = 1 and 0 = 0 are true but never is 1 = 0 or 0 = 1.

1^o =0 ^o

.

1=1

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Sorry but 1=1

I suspect it trickery to do with multiple roots that your father encountered.

e.g. a = sqrt(a) * sqrt(a) , but if you slip in a -ve square root in one of them you get

a = -sqrt(a) * sqrt(a) = -a (a falsehod)

so 2a = a + a = a - a =0

then you can do anything

e.g.

1 = (1+1) / 2

= (1 + (sqrt(1) * sqrt(1)) / 2

= (1 + (1 * -1))/2 (a false step)

= (1 - 1) / 2 = 0

Consider the following problem. Let x=1 and y=1. Then, of course, x=y. Multiplying both sides of the equation by x gives us x2 =xy. Subtracting y2 from both sides makes it x2-y 2 = xy-y2.. If we factor both sides of this equation, we get (x+y)(x-y) = y(x-y). Now, we can divide both sides by (x-y), leaving x+y = y. Substituting the original numbers for x and y produces a simple, yet suspicious result: 2=1.

(X=y=1; then x-y = 0 division by 0 leads to poor result)

To the untrained eye it can be a clever trick, but in reality there's an illegal algebra step, namely, dividing by zero, somewhere in the process of trying to prove 1=0.

A college professor once said you can do lots of stuff if you could divide by zero!

They mostly involve dividing by 0 or some other invalid operation

1 multiplied by 0 = 0 simple wnen you think about it

This can be proved only by using division by zero, which is, of course, a mathematical anomaly, which is what the teacher was demonstrating.