## I'm baffled... help

Suppose that:

Note that this is also true by default:

So we can write previous expression like this:

Shuffle them a little and you get this:

Simplify the expression:

If we omit ( a+b+c ), cause it's legit, we get:

What the duck??? I'm maybe dumb, but I don't see where's the error???

a + b = c

Note that this is also true by default:

4a - 3a = a

So we can write previous expression like this:

4a - 3a + 4b - 3b = 4c - 3c

Shuffle them a little and you get this:

4a + 4b - 4c = 3a + 3b - 3c

Simplify the expression:

4( a+b-c ) = 3( a+b-c )

If we omit ( a+b+c ), cause it's legit, we get:

4=3

What the duck??? I'm maybe dumb, but I don't see where's the error???

I've found one possible solution... but I'll share it after your few tries

uh, I don't ever remember simplifying something like 4(a+b-c) = 3(a+b-c) into 4=3. That doesn't seem right to me. You can however simplify 4(a+b-c) = 3(a+b-c) into a + b - c = 0. which would mean a + b = c.

Your first assumption (eq. 1) a+b=c contains the solution to the apparent impossibility:

If a+b = c then a+b-c = c-c (subtract c from both sides to maintain the equality)

this becomes a+b-c = 0 because c-c = 0

Now substitute this into your final equation;

4(a+b-c) = 3(a+b-c)

4(0) = 3(0)

0 = 0

NOT 4=3

Also, your final term is mistyped (a+b+c) should be (a+b-c)

It is NOT a valid math operation to "omit" a term. You must perform the same mathematical operation (+-*/) on each side of the equation. The is no such operation as "omit"...

If a+b = c then a+b-c = c-c (subtract c from both sides to maintain the equality)

this becomes a+b-c = 0 because c-c = 0

Now substitute this into your final equation;

4(a+b-c) = 3(a+b-c)

4(0) = 3(0)

0 = 0

NOT 4=3

Also, your final term is mistyped (a+b+c) should be (a+b-c)

It is NOT a valid math operation to "omit" a term. You must perform the same mathematical operation (+-*/) on each side of the equation. The is no such operation as "omit"...

I think by "omiting" he meant actually "simplifying", which meant getting rid of (a+b-c)

Getting rid of a+b-c should be allowed. That's what confuses me. Suppose that you change 4 and 3 with x and y. Final statement would be

In this case, if you get rid of a+b-c, it would be perfectly ok, and utterly correct,cause x=y. So, why is it possible with xy, and not with 3 and 4?

x(a+b-c) = y(a+b-c)

In this case, if you get rid of a+b-c, it would be perfectly ok, and utterly correct,cause x=y. So, why is it possible with xy, and not with 3 and 4?

*Getting rid of a+b-c should be allowed. That's what confuses me.*

You're right. It confused me because you simply got rid of it, but as twp said you still have (0) left over, and that makes sense.

I showed this to a friend of mine who studies math and he said that twp was right. You make of this whatever you can.

In math, we don't have an operation named "get rid of". You may (should) use one or more of the four math operators (add, subtract, multiply, divide) IN THAT ORDER. This is known as the order of precedence. Yes, you can and should look this up in any math text.

How do you know that you have an error in your analysis? Try doing an operation OUT OF ORDER.

For instance, try to divide first:

If you wish to ATTEMPT to remove the term a+b-c from both sides of the equation, you could (incorrectly) try to do a divide operation on both sides:

4(a+b-c)/(a+b-c) = 3(a+b-c)/(a+b-c)

However you are then dividing by zero... This gives an undefined result (look it up in any math text).

Attempting to divide by zero is an easy warning flag for any math operation.

The solution is to go back to initial conditions and first reduce them using the four operators (+-*/).

You already know that a+b-c = 0. You may substitute this known relation into the first equation. See my previous post.

4(a+b-c) = 3(a+b-c) where a+b-c = 0

4(0) = 3(0)

0 = 0

The apparent result (4=3) of your original analysis should raise a warning flag that one of your previous assumptions or operations was incorrect (not wrong, just not in the correct order of precedence). So you go back and look at each assumption to find the incorrect step.

What you are attempting to do, in this problem, is known as Simultaneous Solution of Equations. Again, look it up in your math texts.

How do you know that you have an error in your analysis? Try doing an operation OUT OF ORDER.

For instance, try to divide first:

If you wish to ATTEMPT to remove the term a+b-c from both sides of the equation, you could (incorrectly) try to do a divide operation on both sides:

4(a+b-c)/(a+b-c) = 3(a+b-c)/(a+b-c)

However you are then dividing by zero... This gives an undefined result (look it up in any math text).

Attempting to divide by zero is an easy warning flag for any math operation.

The solution is to go back to initial conditions and first reduce them using the four operators (+-*/).

You already know that a+b-c = 0. You may substitute this known relation into the first equation. See my previous post.

4(a+b-c) = 3(a+b-c) where a+b-c = 0

4(0) = 3(0)

0 = 0

The apparent result (4=3) of your original analysis should raise a warning flag that one of your previous assumptions or operations was incorrect (not wrong, just not in the correct order of precedence). So you go back and look at each assumption to find the incorrect step.

What you are attempting to do, in this problem, is known as Simultaneous Solution of Equations. Again, look it up in your math texts.

No need, the important thing is that you understood my question. I wasn't trying to prove that 4=3, that just can't be true.

But seriously now, I didn't knew that if some calculating operation may be possible, doesn't mean it's correct. Thanks

*sound of wormhole opening*But seriously now, I didn't knew that if some calculating operation may be possible, doesn't mean it's correct. Thanks