Answer :
Assume that log 2 is rational, that is,
log2=p/q
where p, q are integers. Since log 1=0 and log10=1,0<log2<1 and p<q.
2=10^p/q
2^p=(2*5)^q
2^q-p=5^p
where q – p is an integer greater than 0.Now, it can be seen that the L.H.S. is even and the R.H.S. is odd. Hence there is contradiction and log 2 is irrational.
log2=p/q
where p, q are integers. Since log 1=0 and log10=1,0<log2<1 and p<q.
2=10^p/q
2^p=(2*5)^q
2^q-p=5^p
where q – p is an integer greater than 0.Now, it can be seen that the L.H.S. is even and the R.H.S. is odd. Hence there is contradiction and log 2 is irrational.
Answer:
[tex]\log 2[/tex] is an irrational number.
Step-by-step explanation:
To show : [tex]\log 2[/tex] is rational or irrational ?
Solution :
We assume that [tex]\log 2[/tex] is a rational number.
So, We can write [tex]\log 2[/tex] in form of p/q where p and q are integers and q is non-zero.
[tex]\log_{10} 2=\frac{p}{q}[/tex]
We know, [tex]\log_b a=x\Rightarrow a=b^x[/tex]
[tex]2=10^{\frac{p}{q}}[/tex]
[tex]2=(2\times 5)^{\frac{p}{q}}[/tex]
[tex]2^q=(2\times 5)^{p}[/tex]
[tex]2^{q-p}=(5)^{p}[/tex]
Where, q-p is an integer greater than zero.
Now, it can be seen that the L.H.S. is even and the R.H.S. is odd.
So, there is contradiction.
As [tex]\log 2[/tex] is an irrational number.