Suppose a, b, c, and d are integers. Prove that if a and b are positive, and a d

Question

Suppose a, b, c, and d are integers. Prove that if a and b are positive, and a divides b, then a lessthanorequalto b. if a and b are positive, a divides b, and b divides a, then a = b. if a divides b, and c divides d, then ac divides bd. If a and b are real numbers, then prove that, |a|/|b| = |a/b| for b notequalto 0. if -b lessthanorequalto a lessthanorequalto b, then |a| lessthanorequalto b. Let a, b and c be integers and x, y, and z be real numbers. Use the technique of backward from the desired conclusion to prove that, if a divides b and a divides b + c, then a divides 3c. if an isosceles triangle has sides of length x, y, and z, where x = y and z = then it is a right triangle.

Explanation / Answer

2)a) given that a,b 0 ; to show a   b

a divides b => b= ax for some +ve integer x => x 1

now, b= ax 1.a = a i.e. b a.

b) a divides b i.e. a|b => ax= b for some +ve integer x (because b>0)

and b divides a i.e. b|a => by= a for some +ve integer y (because a>0)

substituting a from 2nd eqn to 1st eqn, b.y.x=b => y.x=1 => x=y=1

=> a=b.

c) a divides b => b= ax for some integer x - (i)

c divides d => d=cy for some integer y -(ii)

multiplying (i) and (ii)

b.d= a.x.c.y = a.c.x.y [using commutativity]

= (ac)(xy) [using associativity]

=> ac divides bd

3) b) Let -b a b to prove: |a| b

case i : let a 0   

from the inequality a b we get |a| b [ since |a| = a for a   0 ]

case ii: let a < 0

the inequality -b   a => -a   b => |a|   b [ since |a| = -a for a  < 0]

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