What is wrong with this, mathematically?....All numbers are equal.?

Question

Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and let t = a + b. Then

a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b

So all numbers are the same, and math is pointless.

Ok, you guys seem to have a problem. I am going to explain each step in this proof to the best of my ability.

Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and let t = a + b. ....(a, b and t can be any numbers)

Then

a + b = t
(a + b)(a - b) = t(a - b).....multiply both sides by (a - b).
a^2 - b^2 = ta - tb ........multiply out the brackets.
a^2 - ta = b^2 - tb........put terms with 'a' on one side and those with 'b'
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4......add "(t^2)/4" on both sides
(a - t/2)^2 = (b - t/2)^2........factorise both sides.......find the factors.
a - t/2 = b - t/2...........Take the positive square root on both sides.
a = b..........add t/2 on both sides, and you are left with this result.

Noting that a, b and t can be any numbers; then all numbers are equal.

Answer 1

In going from

(a - t/2)^2 = (b - t/2)^2

to

a - t/2 = b - t/2

You assumed that a - t/2 >=0 and b - t/2 >=0

However, form a+b = t and a - t/2>=0,
subtracting we get b <= t/2,
your assumption that b - t/2 >= 0, gives b >= t/2.
All this forces b = t/2.
Further substitution gives a = t/2.
Your assumptions are equivalent to a = t/2 = b.

Since you assumed it, it comes as a result.

Answer 2

Your evidence is high quality until eventually the subsequent-to-very last step, the position you're taking the sq. root. sqrt(a*a) = +/-a, no longer only a. at the same time as it really is real that (-3)^2 = (3)^2 this would not propose that 3 = -3 note that t/2 is largely the overall of a and b. So, (a-t/2) and (b-t/2) have an identical importance, yet opposite indicators. The squares are equivalent, yet that would not propose the numbers are equivalent.

Answer 3

The step sqrt[(a-t/2)^2] = a-t/2 is only partially correct. Remember that there are 2 answers to a sqrt. The other answer is -a+t/2. If you rewrite the next to last equation as -a+t/2 = b-t/2 and then solve :
-a-b = -t/2 - t/2 you get -a-b= -t
multiply by -1 and you have a+b = t
Nothing was proved.

Answer 4

the reason is square root of a^2 is both + a and -a and they are not same

(a-t/2)^2 = (b-t/2)^2

means a-t/2 = b- t/2 or -(b-t/2)

here second one holds

Answer 5

When you muliply both sides by (a-b), this would be true only if a does not equal b. Why? because otherwise you will be multiplying both sides by zero.

Answer 6

OK, have it your way. Good luck convincing your bank, though, or your boss.

Answer 7

all the numbers are the same!