Help in Pseudo-Complex numbers!

[prhilado]

A pseudo-complex number is a number of the form a + bt where a, b are real num- bers and t is a symbol such that t^2 = 1. Multiplication is defined by the law
(a+bt)(c+dt):=(ac+bd)+(ad+bc)t

---

Does the pseudo-complex number 3 + t have a multiplicative inverse? If so, find it; if not, give a proof.

I tried a method where I let u = 3+t and supposed u had a multiplicative inverse v. I then let v*(3+t) = 1 (by the rules of multiplicative inverse). I then tried to manipulate the equation in hopes of arriving at some sort of contradiction, and came up with nothing. Please help!

Many thanks!

 

[pka]

Does the pseudo-complex number 3 + t have a multiplicative inverse? If so, find it; if not, give a proof.

What in the world is a Pseudo-Complex number ?

 

[prhilado]

What in the world is a Pseudo-Complex number ?

According to my notes, "A pseudo-complex number is a number of the form a + bt where a, b are real numbers and t is a symbol such that t^2 = 1. Multiplication is defined by the law(a+bt)(c+dt):=(ac+bd)+(ad+bc)t"

 

[daon2]

According to my notes, "A pseudo-complex number is a number of the form a + bt where a, b are real numbers and t is a symbol such that t^2 = 1. Multiplication is defined by the law(a+bt)(c+dt):=(ac+bd)+(ad+bc)t"

Assuming "1" is equal to 1+0t (you haven't defined it), solve the system:

3c+d=1
3d+c=0

 

[Dale10101]

Two cents

Assuming "1" is equal to 1+0t (you haven't defined it), solve the system:

3c+d=1
3d+c=0

First problem, find the identity element. The identity element is the element that does nothing. That is, multiplying by 1 returns just what you put in. So,

(a+bt)(m+nt) = (a+bt), what is m, n that makes that happen?

(a+bt)(m+nt) =(am+bn)+(an+bm)t = (a+bt)

By eyeball m must be 1, and n must be 0, so the identity element is (m+nt) = (1+0t)

The next problem, find the multiplicative inverse. A number multiplied by its multiplicative inverse always return the identity element. Real number example: 24(1/24) = 1, (1/24) is the multiplicative inverse of 24.

So, if (3+t) has a multiplicative inverse then it must be a solution to the following equation:

(3+t)(c+dt) = (1+0t),

where (c+dt) is the hypothesized multiplicative inverse of (3+t), and (1+0t) is, of course, the identity element.

Carrying out the multiplication gives you M. Daon2's equations and his question .... which, thus, by contributing nothing new makes me an identity element ... wait, is that good thing?