Mathematical proof for 2 = 3

[mandala]

Came across an equation that seeks to prove 3 = 2, and turn our world upside down. Can someone explain, at a high-school algebra level, why this might be incorrect?

Here are the equations:

-6 = -6

So, 9-15 = 4-10

adding 25/4 to both sides:

9-15+25/4 = 4-10+25/4

Changing the order:

9+25/4-15 = 4+25/4-10

This is of the form a square + b square - 2.a.b = (a-b) square.

So this equation can be written as:
(3-5/2)(3-5/2) = (2-5/2)(2-5/2)

Taking positive square root on both sides:
3 - 5/2 = 2 - 5/2

So, 3 = 2.

Thoughts?

[Carmady]

@mandala

Taking positive square root on both sides:
3 - 5/2 = 2 - 5/2

(bolding mine)

[Sunspace]

Somebody divided by zero! I’m telling!

[Jragon]

The trick is that you can’t take that square root without evaluating the ^2 first.

Normally you can say sqrt(n^2) = 5 where “n” is some expression because they’re functionally equivalent. The trick here is that “n” is a negative number in one case.

4+25/4-10 = -1/2

9+25/4-15 = 1/2

So

sqrt(1/2^2) = sqrt((-1/2)^2)

=> sqrt(1/4) = sqrt(1/4)
=> 1/2 = 1/2

[John_DiFool]

Thanks-I knew it was an order of operations error of some sort, that would do it.

[Jragon]

@Jragon

Normally you can say sqrt(n^2) = 5 where “n” is some expression because they’re functionally equivalent. The trick here is that “n” is a negative number in one case.

Uh… sqrt(n^2)=n

Actually, nevermind. Every square root is equal to five, screw it, it makes everything easier.

[leahcim]

@Jragon

Uh… sqrt(n^2)=n

No, it’s not. At least if you take “sqrt” to mean the positive square root.

sqrt(n^2) = n for n >= 0, and sqrt(n^2) = -n for n < 0.

The OP’s proof fails because when it gets to (3-5/2)(3-5/2) = (2-5/2)(2-5/2), he is saying:

(1/2) ^2 = (-1/2) ^2

Which is entirely true, but then he concludes, through careless application of a positive square root that:

1/2 = -1/2

Which is false.

[Jragon]

@leahcim

No, it’s not. At least if you take “sqrt” to mean the positive square root.

sqrt(n^2) = n for n >= 0, and sqrt(n^2) = -n for n < 0.

The OP’s proof fails because when it gets to (3-5/2)(3-5/2) = (2-5/2)(2-5/2), he is saying:

(1/2) ^2 = (-1/2) ^2

Which is entirely true, but then he concludes, through careless application of a positive square root that:

1/2 = -1/2

Which is false.

Which is exactly what I explained in my first post, I was correcting the phrase “we can usually say sqrt(n^2)=5” where I meant “sqrt(n^2)=n”. I then went on the explain exactly what you explained – that in this case it doesn’t work because “n” is negative, so we can’t use the same sqrt(n^2)=n identity that we use when n >=0.

[Asympotically_fat]

2 and 3 turn would turn out to be the same number all along if arithmetic had been axiomised by M Night Shyamalan.

[Jragon]

@Sunspace

Somebody divided by zero! I’m telling!

I’m actually surprised it wasn’t division by zero. These kinds of proofs always involve dividing by zero. The mathtrolls are getting smarter.

[leahcim]

@Jragon

Which is exactly what I explained in my first post, I was correcting the phrase “we can usually say sqrt(n^2)=5” where I meant “sqrt(n^2)=n”. I then went on the explain exactly what you explained – that in this case it doesn’t work because “n” is negative, so we can’t use the same sqrt(n^2)=n identity that we use when n >=0.

Sorry, I mis-parsed and thought that you were actually arguing that sqrt(n^2) = n in all cases in the post I was replying to, rather than making a correction to your first post.

[Find_Friends]

@Asympotically_fat

2 and 3 turn would turn out to be the same number all along if arithmetic had been axiomised by M Night Shyamalan.

It’s not only tricks involving division by zero and tricks involving sqare roots that have been used to equate all numbers and destroy arithmetic. Basic differential calculus can be used to “prove” 1=2 and thus all numbers are equal to each other. The “proof” has few steps, is easy to read, and is very straight-forward. Or so it seems.

Will I hear crickets chirping or will someone here beg me to show it? One math enthusiast I showed it to described it as “delicious” and only half of his instuctors spooted the mis-step.

[Leo_Bloom]

tl;dr

Just caught this thread. OP’s wrong, right? Otherwise I gotta start packing because something’s gonna blow…

[Roland_Orzabal]

@Trans Fat Og

It’s not only tricks involving division by zero and tricks involving sqare roots that have been used to equate all numbers and destroy arithmetic. Basic differential calculus can be used to “prove” 1=2 and thus all numbers are equal to each other. The “proof” has few steps, is easy to read, and is very straight-forward. Or so it seems.

Will I hear crickets chirping or will someone here beg me to show it? One math enthusiast I showed it to described it as “delicious” and only half of his instuctors spooted the mis-step.

Never seen a calculus version, so sure, post away. If nothing else I can use it to screw with the Calc 1 kids I tutor.

[Asympotically_fat]

@Trans Fat Og

It’s not only tricks involving division by zero and tricks involving sqare roots that have been used to equate all numbers and destroy arithmetic. Basic differential calculus can be used to “prove” 1=2 and thus all numbers are equal to each other. The “proof” has few steps, is easy to read, and is very straight-forward. Or so it seems.

Will I hear crickets chirping or will someone here beg me to show it? One math enthusiast I showed it to described it as “delicious” and only half of his instuctors spooted the mis-step.

Yep, there’s a lot of 2=1 ‘proofs’, I guess the one you may be thinking of is (IIRC):

x[sup]2[/sup] = x*x

x*n = x+…+x with the number of x’s on the RHS being n

Therefore x[sup]2[/sup] = x+…+x with x number x’s

Setting f(x) = x[sup]2[/sup]

and g(x) = x+…+x (with x number of x’s)

Then

f(x)=g(x)

and therefore

f’(x)=g’(x)

From basic calculus:

f’(x)=2x

and

g’(x) = 1+…+1 (with x number of 1’s) = x

Therefore:

2x=x

Divide by x:

2=1

QED

Of course it’s fairly obvious where that one goes wrong.

[Roland_Orzabal]

Oops, misread the setup. Ignore this post. Will follow up later. :smack:

[Asympotically_fat]

@Roland_Orzabal

Oops, misread the setup. Ignore this post. Will follow up later. :smack:

Well you were right though, it isn’t particularly sophisticated. I think I may’ve seen a much better one involving calculus, but if I did I can’t for the life of me remember what it was.

[Jragon]

I… uh… I can’t find the error. The best I can say is that the setup of g only really works if x is integral, but that’s kind of a weak refutation since it shouldn’t even work if x is integral. The other issue is that x might be 0, but that doesn’t feel like a magic bullet either.

I feel dumb.

[SCAdian]

@Jragon

The trick is that you can’t take that square root without evaluating the ^2 first.

Normally you can say sqrt(n^2) = 5 where “n” is some expression because they’re functionally equivalent. The trick here is that “n” is a negative number in one case.

4+25/4-10 = -1/2

9+25/4-15 = 1/2

Huh?

25/4 = 6.25

4+25/4-10 = 4+6.25-10 = 10.25-10 = .25

9+25/4-15 = 9+6.25-15 = 15.25-15 = .25

[leahcim]

@Jragon

I’m actually surprised it wasn’t division by zero. These kinds of proofs always involve dividing by zero. The mathtrolls are getting smarter.

You can always convert a wrong proof that assumes a^2 = b^2 -> a = b to one that wrongly assumes that xa = xb -> a = b, even if x is 0, via the process:

a^2 = b^2 -> a^2 - b^2 = 0 -> (a + b)(a - b) = 0 -> a - b = 0 -> a = b

I wonder if there is a way to make the reverse transformation to have a full “equivalence of wrong proofs” theorem.