Question

any body can find the mistake in the following let a=b 2a=2b
(7-5)a=(7-5)b 7a-5a = 7b-5b 7(a-b) = 5 (a-b) Therefore 7=5

Answer 1

Everything's right, except the process in simplifying 7(a-b)=5(a-b)

Yes, we know that $a\times c=b\times c \implies a = b$

But this holds only when for $c\neq 0$

Why?

Let us examine the equation

$a\times c=b\times c$

We know that the multiplicative inverse of 0 is not defined i.e., $c^{-1}$ is not defined for $c=0$

as there is no number n such that $n\times0=1$

So, If $c\neq 0, \:c^{-1}$ exists

Multiply both sides by the $c^{-1}$$(a\times c)\times c^{-1}=(b\times c)\times c^{-1}\\\implies a\times (c\times c^{-1})=b\times (c\times c^{-1}) \:\:\:[Distributive\:property\:of\:multiplication]\\\implies a = b\:\:\:\:for\:c\neq 0$

Here,

7(a-b)=5(a-b)

In the question, the inverse of (a-b) is multiplied to both sides to get 7=5

But a=b, so a-b=0 implying that a-b has no multiplicative inverse

∴The process of cancelling a-b from both sides makes no more sense as it involves something that doesn't even exist

So, we continue the process of simplifying by substitution

7(a-b)=5(a-b)

⇒7×0=5×0

⇒0=0

[⇒7(a-b)=5(a-b) is a tautology]

This process takes a different route away from resulting in 7=5 and ends up implying only one fundamental thing (0=0), irrelevant of tautologies of the statement.