Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 117116 by mathocean1 last updated on 10/Oct/20

x , y ,z , t ∈ Z.  x and y are x are respectively the  divisor of y and t.  Justify the existence of k ∈ Z such  that yt=xzk.  Deduct that x^(m ) is a divisor of y^m   where m ∈ N.

$${x}\:,\:{y}\:,{z}\:,\:{t}\:\in\:\mathbb{Z}. \\ $$$${x}\:{and}\:{y}\:{are}\:{x}\:{are}\:{respectively}\:{the} \\ $$$${divisor}\:{of}\:{y}\:{and}\:{t}. \\ $$$${Justify}\:{the}\:{existence}\:{of}\:{k}\:\in\:\mathbb{Z}\:{such} \\ $$$${that}\:{yt}={xzk}. \\ $$$${Deduct}\:{that}\:{x}^{{m}\:} {is}\:{a}\:{divisor}\:{of}\:{y}^{{m}} \\ $$$${where}\:{m}\:\in\:\mathbb{N}. \\ $$

Commented by 1549442205PVT last updated on 10/Oct/20

In R there is  no  concept of  a  divisor  of a number.It only is said in Z.You  look at question again,please.

$$\mathrm{In}\:\mathrm{R}\:\mathrm{there}\:\mathrm{is}\:\:\mathrm{no}\:\:\mathrm{concept}\:\mathrm{of}\:\:\mathrm{a}\:\:\mathrm{divisor} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{number}.\mathrm{It}\:\mathrm{only}\:\mathrm{is}\:\mathrm{said}\:\mathrm{in}\:\mathrm{Z}.\mathrm{You} \\ $$$$\mathrm{look}\:\mathrm{at}\:\mathrm{question}\:\mathrm{again},\mathrm{please}. \\ $$

Commented by mathocean1 last updated on 10/Oct/20

exactly it was an error

$${exactly}\:{it}\:{was}\:{an}\:{error} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com