All numbers are equal?

Question

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

Therefore all numbers are equal.

Where did this equation go wrong?

Answer 1

A couple places...

At step 1, if a and b were equal, then you are multiplying both sides by 0.

Also, at the penultimate step, when you take the square root of both sides you need need to account for plus and minus. You really should have

a - t/2 = ±(b - t/2)

The positive case leads back the contradiction in step 1 (multiplying by 0).

The negative case becomes:
a - t/2 = -b + t/2

Adding b to both sides and t/2 to both sides you get:
a + b = t

Back to square one!

Answer 2

This part is good, that is, each equation follows from those above for all real numbers:
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

But a - t/2 = b - t/2
does NOT follow (for example, let a=1 b=3).
What DOES follow is that
|a - t/2| = |b - t/2|
i.e. a and b are the same distance from t/2, which is, after all the average of a and b.

Answer 3

On the third line you didn't not do the same thing to each side. All you did was switch ta and b^2. If you - ta from each side it's a^2-B^2-ta=-tb or you could do +b^2 which then would be a^2=ta-tb+b^2 so maybe that's wrong...

Answer 4

When you took the square root of (a - t/2)^2 = (b - t/2)^2
you did not consider that one side may be minus the other:
x^2 = (-x)^2

lets examine the second to last row and you'll see that that is exactly what happened:
a - t/2 = b - t/2
a - (a+b)/2 = b -(b+a)/2
a/2 -b/2 = b/2 -a/2
a/2 -b/2 = -(a/2 -b/2)

Answer 5

You did not consider principal square roots.

If you start with 6+4=10, using a=6, b=4 and t=10....

I believe when you get down to the sixth line of math, you get one (six minus five) squared is equal to negative one (four minus five) squared.

You then took the square root of both sides, which could mean that six minus five is equal to four minus five.

I don't think sooooo....

Imagine your last line of math... placing all of your t terms on the right... it would be:
a-b=0

Your completing the square step would only work if a would be equal to b, thus the difference between the two would be zero, so t would equal zero.

Remember, the square root of a-squared is the absolute value of a... plus, if they were in fact equal, then your second line would give us zero equals zero in the third line....

Answer 6

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

all this is good

the rest is bull

Answer 7

When squaring the additive inverse of a number, the result is the same as the square of the original number.

a - t/2 = b - t/2

BUT! It could be that a - t/2 is equal to the additive inverse of (b - t/2).

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

Answer 8

The equation is not true for all numbers. it is not an Identity. As Already pointed out, the positive root of the quadratic shows that all you are really proving 0 = 0.

Answer 9

You eliminated the "T" term. That means, T is no longer a part of your proof. T could have been any number. Therefore, you can not say a=b=t.

Also, when you took square_root of both sides, you forgot to consider, if a^2 = b^2 then a=+/- b.

Answer 10

when you go from line 5 to line 6
you put ta = tb to cancel them (you did not write them in line 6)
You did not say that ta = tb before
therefor this equation is wrong
a is not equal to b