We all know that 0 can never be equal to 1, and 1 can never be equal to 2. But using some tricky ways we can prove 0=1 and 1=2. And of course, there must be some mistakes in the proofs. Let’s find out below.

Table of Contents

**0=1 Proof**

**First Proof of 0=1: **To conclude 1=2, we will use an infinite series. See the below steps. You need to spot the mistakes in the proof.

**Step 1:** We have

0=0+0+0+…

**Step 2:** Write each zero on the right-hand side as 0=1-1. So we get that

0=(1-1)+(1-1)+(1-1)+…

**Step 3:** Rearrange the parentheses in the following way:

0=1+(-1+1)+(-1+1)+(-1+1)+…

**Step 4:** Put -1+1=0

0=1+0+0+0+…

So finally we get that

**0=1**

**[Second Proof of 0=1]**

**Step 1:** We have

0=0

**Step 2:** Rewriting it we obtain that

12-12=16-16

⇒ 3.4-3.4 = 4.4-4.4

**Step 3:** Taking commons we have

3(4-4)=4(4-4)

**Step 4:** Cancelling 4-4 from both sides, we get that

$3 \cdot \cancel{(4-4)}=4 \cdot \cancel{(4-4)}$

$\Rightarrow 3=4$

$\Rightarrow 3+0=3+1$

Thus, we have

**0=1**

**[Third Proof of 0=1]**

**Step 1:** We have

-20=-20

**Step 2:** Rewriting -20 of both sides we have

16-36=25-45

⇒ 4^{2}-4.9 = 5^{2}-5.9

**Step 3:** Adding 81/4 to both sides, we obtain

$4^2-4\cdot 9+\frac{81}{4}$ $=5^2-5\cdot 9 +\frac{81}{4}$

$\Rightarrow 4^2-2\cdot 2\cdot \frac{9}{2}+(\frac{9}{2})^2$ $=5^2-2\cdot 5\cdot \frac{9}{2}+(\frac{9}{2})^2$

$\Rightarrow (4-\frac{9}{2})^2=(5-\frac{9}{2})^2$

**Step 4:** Taking square roots on both sides, we have

$4-\frac{9}{2}=5-\frac{9}{2}$

**Step 5:** Cancelling 9/2 from both sides, we have

$4-\cancel{\frac{9}{2}}=5-\cancel{\frac{9}{2}}$

$\Rightarrow 4=5$

$\Rightarrow 4+0=4+1$

It follows that

**0=1**

**1=2 Proof**

**First Proof of 1=2: **In the first method, we will use algebra. We will follow the following steps to conclude 1=2. But this is not true. So there must be a mistake in the proof. You have to find them.

**Step 1:** Let a=b

**Step 2:** Multiplying both sides by a, we get

a^{2}=ab

**Step 3:** Subtract b^{2 }from both sides.

a^{2}-b^{2}=ab-b^{2}

**Step 4:** Factorise both sides. Doing that we get

(a-b)(a+b)=b(a-b)

**Step 5:** Cancelling a-b from both sides, we have

a+b=b

**Step 6:** Put b in the place of a as we have a=b.

b+b=b

⇒ 2b=b

⇒ **2=1**

**Second Proof of 1=2: **To prove 1=2, we will now use the theory of calculus. First, observe the following patterns:

2^{2}=2+2 (2 times)

3^{2}=3+3+3 (3 times)

4^{2}=4+4+4+4 (4 times)

$\vdots \quad \vdots \quad \vdots$

$x^2=x+x+\cdots +x$ (x times)

Differentiating both sides with respect to x, we get

$\dfrac{d}{dx}(x^2)$ $=\dfrac{d}{dx}(x)+\dfrac{d}{dx}(x)+\cdots +\dfrac{d}{dx}(x)$

$\Rightarrow 2x=1+1+\cdots +1$ (x times)

$\Rightarrow 2x=x$

$\Rightarrow 2=1$