If w \alpha = ( \alpha |x \: e \: w) \: then \: w3 \: n \: w5 = option 1) W3 2) W5 3) W8 4)

Question

If
w \alpha  = ( \alpha  |x \: e \: w) \: then \: w3 \: n \: w5 =
option 1) W3 2) W5
3) W8 4) W15
Don’t write only answer give explanation I know correct answer is 4) W15 But How ?? ​

in progress 0
Gianna 2 weeks 2021-10-01T21:31:19+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-10-01T21:32:42+00:00

    Correct Question:-

    If \sf{W_{\alpha}=\{x:\alpha\mid x,\ x\in\mathbb{W}\},} then \sf{W_3\cap W_5=}

    \sf{1.\quad W_3}

    \sf{2.\quad W_5}

    \sf{3.\quad W_8}

    \sf{4.\quad W_{15}}

    Solution:-

    By definition,

    \longrightarrow\sf{W_3=\{x:3\mid x,\ x\in\mathbb{W}\}}

    And,

    \longrightarrow\sf{W_5=\{x:5\mid x,\ x\in\mathbb{W}\}}

    For two non – empty sets \sf{A} and \sf{B} we know that,

    \longrightarrow\sf{A\cap B=\{x:x\in A\ and\ x\in B\}}

    Hence,

    \longrightarrow\sf{W_3\cap W_5=\{x:3\mid x\ and\ 5\mid x,\ x\in\mathbb{W}\}\quad\quad\dots(1)}

    By divisibility rule, if a number is exactly divisible by 3 as well as 5, then it is exactly divisible by 15 too.

    \longrightarrow\sf{3\mid x\ and\ 5\mid x\implies15\mid x}

    Therefore (1) becomes,

    \longrightarrow\sf{W_3\cap W_5=\{x:15\mid x,\ x\in\mathbb{W}\}}

    By definition,

    \longrightarrow\underline{\underline{\sf{W_3\cap W_5=W_{15}}}}

    Hence 4th option is the answer.

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )