Mathematical Fallacy 2: Is $3458=1729$ ?

Algebra Level 1

$\begin{aligned} 1729^2=&1729 \cdot 1729 \quad & \ldots (1) \\ 1729^2-1729^2=&1729 \cdot 1729-1729^2 \quad & \ldots (2) \\ (1729+1729)(1729-1729)=& 1729(1729-1729) \quad & \ldots (3) \\ (1729+1729)\cdot \cancel{0}=&1729\cdot \cancel{0} \quad & \ldots (4) \\ 3458=&1729 \end{aligned}$

The above shows my attempt to prove that $3458 = 1729$ . In which of these steps did I make a flaw in my logic?

1 2 3 4

5 solutions

Rishabh Jain
Feb 22, 2016

Let $a,b$ and $c$ be real numbers, then according to properties of real numbers: $\Large ab=ac\implies b=c$ is justified only when $a\neq 0$ since to obtain $b=c$ from previous step we have to multiply by $\frac{1}{a}$ on both sides and division by 0 is not defined. Hence mistake is commited in $step~\boxed 4$ .

(1729+1729)(1729-1729)=1729(1729-1729) but if (1729+1729)(1729-1729)/(1729-1729)=1729 then (1729+1729)(1)=1729 ?????????????????

Owais Saeed - 5 years, 3 months ago

Satyabrata Dash
Feb 29, 2016

0/0 is undefined . Thus,we see that the 4th step is incorrect.

Rohit Udaiwal
Feb 22, 2016

In the 4th step, we have cancelled $0$ but this is wrong since cancellation by zero is undefined.

This is just the 1=0 "proof", isnt it?

Teng Chou Nathan Mar - 5 years, 3 months ago

Curtis Duong
Jun 8, 2020

At step 4, you divide by zero.

Kevin Silva
Feb 29, 2016

Any number multiplied by 0 is always 0

Mathematical Fallacy 2: Is 3458 = 1729 3458=1729 3 4 5 8 = 1 7 2 9 ?

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Mathematical Fallacy 2: Is $3458=1729$ ?