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anonymous
 one year ago
Are any of the following statements is correct :
i) 2  a ⇒ 2 a^2
ii) 2  a^2 ⇒ 2  a
iii) 2  a ⇔ 2  a^2
for integers a ?
anonymous
 one year ago
Are any of the following statements is correct : i) 2  a ⇒ 2 a^2 ii) 2  a^2 ⇒ 2  a iii) 2  a ⇔ 2  a^2 for integers a ?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you think a statement is false, try to produce a counterexample.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jayzdd but how do I show it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like in the first one\[a = 2*x \rightarrow a*a=2*x\] and because 2 is a divider to a it must be a divider to a*a right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@nincompoop that equal sign is if I should prime factorize a, and we know that 2 is a divider of a?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@nincompoop could you explain instead?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1(2n+1)(2n+1) = 4n^2+4n+1 = 2(2n^2+2n) + 1 = 2k+1 an odd squared, is odd but then since 2 doesnt divide odd numbers, it doesnt divide either factor to start with even*odd, divides the even number, and therefore the product and its square.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@amistre64 could you explain a little more what your doing up there?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1just working some thoughts ...  spose we have an even number, it is divisible by 2 if we multiply it by itself, we have 2 factors, each divisible by 2  spose we have an odd number, it is NOT divisible by 2 if we multiply it by itself, we have 2 factors, NEITHER are divisible by 2  spose we have a number that is the product of an even and an odd, it is divisible by 2 if we multiply it by itself, we have 2 factors that are still divisible by 2  ergo if a number is divisible by 2, its square is divisible by 2 if a squared number is divisible by 2, then its primary value is also divisible by 2 this creates a bidirectional setup

amistre64
 one year ago
Best ResponseYou've already chosen the best response.12n(2n) = 2(2n^2) 2n(2m+1) = 2(2nmk+n) = 2k, an even number odd number are not divisible by 2, so any combination of odd numbers will not be divisible by 2 to start with.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i) 2  a ⇒ 2 a^2 if a is even, then a^2 is even .... so true ii) 2  a^2 ⇒ 2  a since a^2 is even only if a is even ... this is also true iii) 2  a ⇔ 2  a^2 since i and ii are true, iii is correct by default.
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