# Can 1 = 2 ?

1. Sep 4, 2006

### navneet1990

can 1 = 2 ??

is this possible
1 = 2
??

2. Sep 4, 2006

### waht

Consider the equation,

x + 1 = x + 2

And you are asked to solve for x.

By looking at it, you can say there is no solutions.

Subtract x from both sides, and you get

1 = 2

But what if you could subtitute 1 = 2 in one eqation

you get x + 1 = x + 1

or x + 2 = x + 2

That makes more sense now.

3. Sep 4, 2006

### CRGreathouse

Sure, if the underlying assumptions are inconsistent.

4. Sep 4, 2006

### navneet1990

i didnt get it
for any value of x how can x+1 = X+2
the assumption itself is wrong
right??

5. Sep 4, 2006

This really has nothing to do with math.

6. Sep 4, 2006

### arildno

Sure 1=2 in some number systems.
In a number system that denies the axiom $0\neq{1}$, for example, 1=2 will be a true statement.

7. Sep 4, 2006

### StatusX

1 and 2 are just symbols. By their standard definition, they correspond to two distinct real numbers, and so the statement '1=2' is false. If you want to use these symbols in a different way, then the statement '1=2' can be true, false, or meaningless, depending on how you define them.

8. Sep 5, 2006

### loom91

1 is not equal to 2 in any of the standard formulations of natural numbers (Peano axioms, set theory etc). However you can very easily define two symbols 1 and 2 in a number system such that they are equal.

9. Sep 5, 2006

### navneet1990

cant we use complex numbers to prove it
like :

i = i

root -1 = root -1

hence ,

root -1/ root 1 = root -1 / root 1

hence,
whole root [-1/1] = whole root [ -1/1]

hence,
whole root [-1/1] = whole root [ 1/-1]

[ -5/4 can also be written as 5/-4...cant it??? i mean -1/1 is the same as 1/-1 right??]

hence,
root -1 / root 1 = root 1 / root -1

hence,

i / 1 = 1 / i

i square = 1

hence,
-1 = 1

3/2 + ( -1 ) = 3/2 + 1

3/2 - 1 = 3/2 + 1

2/ 2 = 4/2

therefore,

1 = 2

can this be possible

10. Sep 5, 2006

### J77

Nope.

You can't take roots on both the top and bottom like that.

$$i=e^{i\pi/2}$$

$$1/i=e^{-i\pi/2}$$

With these 1=2 things, there's always a mistake/trick.

11. Sep 5, 2006

### navneet1990

um...
i kinda understood a little
but i didnt understand
the

i = e raised to (-i Pie/ 2)
what is that

12. Sep 5, 2006

### chroot

Staff Emeritus
He's expressing complex numbers as complex exponentials.

$$i=e^{i\pi/2}$$

is the value on the complex unit circle corresponding to an angle of pi/2. It is equal to i.

His point is that 1/i and i/1 are quite different numbers, on opposite sides of the unit circle, so your "proof" contains an error.

- Warren

13. Sep 5, 2006

### navneet1990

ohk
thank you

14. Sep 5, 2006

A rational number is, by definition, a number of the form $$\frac{p}{q}$$, where $$p \in Z$$ and $$q \in N$$. So, you can't write -5/4 as 5/-4.

15. Sep 5, 2006

### arildno

Eeeh, wherever do you have this limitation from??
Not saying you might not be right, but I really don't see the necessity of this limitation.

16. Sep 5, 2006

### jnorman

i remember my father showing me a proof once that 1=2, but cant quite recall it. but, if you start with the equation:
x^2 -1 = 0, you can factor x^2 - 1 into (x+1)(x-1)=0
then divide both sides by x-1, and get x+1=0.
for a value of x=1, you have shown that 2=0.
:-)

17. Sep 5, 2006

### chroot

Staff Emeritus
You should tell your father to take more math classes. You cannot divide both sides by x-1, when x=1, because that is equivalent to division by zero. Division by zero is not a "legal" mathematical operation.

- Warren

18. Sep 5, 2006

### mattmns

That is NOT the definition of a rational number

http://mathworld.wolfram.com/RationalNumber.html

A rational number is a number p/q where p and q are Integers and $q \neq 0$

So, you can write (-5)/4 as 5/(-4)

19. Sep 5, 2006

### CRGreathouse

How quaint.

20. Sep 5, 2006

### HallsofIvy

A rational number is, by definition, a number which can be written in the form $\frac{p}{q}$ where $p \in Z$ and $q \in N$. Whether or not the number is written that way is irrelevant.

Yes, you can write -5/4 as 5/-4 just as you could write it as -1.25.

21. Sep 5, 2006

### tony873004

1 quart = 2 pints

22. Sep 5, 2006

It's a standard definition. Btw, does it appear logical to divide with a negative number as well as take Wolfram definitions sooo seriously ..Or let's state it this way: there is no need for q to be an integer. It is enough for q to be a natural number.

Last edited: Sep 5, 2006
23. Sep 5, 2006

### shmoe

Allowing any non-zero integer in the denominator is much more standad from books I've seen. This is what happens in the more general case of the field of fractions of an integral domain where you don't necessarily have any sort of order.

Yes it does. Why wouldn't it be?

Wolfram is generally pretty good at reflecting the 'usual' definitions. This one fits with my experience. You could define them as Halls says, as things that *can* be written a/b where a,b are integers and b>0, but any definition that tries to say 5/(-4) is not rational, and is not the same as (-5)/4 is non-standard and different from everyone else's.

24. Sep 6, 2006

### arildno

So, you didn't have any arguments after all.
Of course 1/(-4) has a neat interpretation:
It is the number that multiplied with (-4) equals 1.
The number (-1)/4 is, by definition equal to (-1)*(1/4).

25. Sep 6, 2006

Right, but let's put it this way. Let's define the set of rational numbers as $$Q=\left\{\frac{p}{q}:p \in Z , q \in N\right\}$$. If we compare two different rational numbers, then we have $$p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}>\frac{p_{2}}{q}$$. Now, let's define the set of rational numbers as $$Q=\left\{\frac{p}{q}:p, q \in Z , q \neq 0 \right\}$$. Then we have $$p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}>\frac{p_{2}}{q}$$ if $$q > 0$$ and $$p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}<\frac{p_{2}}{q}$$ if $$q < 0$$. So, in the first case, it's easier to compare two rational numbers, which may make the first definition more convenient. Sorry if I'm tiresome, ( ) but it's the defiition that I found in almost all my math textbooks (mathematical analysis, elementary math, etc.), so I'm convinced there's a reason for it.