Question

log2 rational or irrational? Justify your answer.

Answer 1

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. 

Answer 2

Answer:

$\log 2$ is an irrational number.

Step-by-step explanation:

To show : $\log 2$ is rational or irrational ?

Solution :

We assume that $\log 2$ is a rational number.

So, We can write $\log 2$ in form of p/q where p and q are integers and q is non-zero.

$\log_{10} 2=\frac{p}{q}$

We know, $\log_b a=x\Rightarrow a=b^x$

$2=10^{\frac{p}{q}}$

$2=(2\times 5)^{\frac{p}{q}}$

$2^q=(2\times 5)^{p}$

$2^{q-p}=(5)^{p}$

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 $\log 2$ is an irrational number.