Question

Show that ab=ba, where a and b are both real.

Answer 1

Suppose, \(ab \ne ba\)
then, \(ab - ba\) must equal some non zero real number \(k\)

\(ab - ba = k\)

divide both sides by \(a\)

\(\frac{ab - ba}{a} = \frac{k}{a}\)

\(\frac{ab}{a} - \frac{ba}{a} = \frac{k}{a}\) ***** is this ok ?

\(b-b = \frac{k}{a}\)

\(0 = \frac{k}{a}\)

\(0 = k\)

contradiction

Answer 2

**** looks ive used distributive law or something in that line, im not sure if its okay to use...

Answer 3

@rsadhvika
No its not true, sorry.

Answer 4

i thought so, u mean we cant use distrbitive law right ?

Answer 5

You have used the theorem throughout the solution.
4th and 5th parts aren't true. because we never known the equality 1b-b1=b-b .....

Answer 6

ohk i see that

Answer 7

ive used 1b = b1 = b which i should not use

Answer 8

Actually.

Answer 9

let a =1 b = 3
therefore ab=1*3=3
ba=3*1=3
but ab is also=3
therefore ab=ba

Answer 10

@uditkulka it's wrong.........

Answer 11

why?

Answer 12

Because you've used the axiom throughout the solution! 3*1=1*3 ????