log2 rational or irrational? Justify your answer.
Answers
Answered by
83
let us assume log 2 as rational,that is
log 2=p/q -------1
where p,q are integers.
Since log 1 =0 and log 10=1,0<log 2<1 and therefore p<q.
from 1, 2=10powerp/q
2powerq=(2*5)power p
2power q-p= 5 power p, where q-p is an interger greater than 0.
Now it can be seen that L.H.S is even and the RHS is odd.Hence there is contradiction and therefore log2 is irrational.
log 2=p/q -------1
where p,q are integers.
Since log 1 =0 and log 10=1,0<log 2<1 and therefore p<q.
from 1, 2=10powerp/q
2powerq=(2*5)power p
2power q-p= 5 power p, where q-p is an interger greater than 0.
Now it can be seen that L.H.S is even and the RHS is odd.Hence there is contradiction and therefore log2 is irrational.
Answered by
43
Answer:
Irrational
Step-by-step explanation:
Let, log 2 base 10 = x
2=10 power x
This not true for any rational value of x
Therefore log 2 base 10 is an irrational number
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