Question

prove that $\sqrt7$ is an irration number

Answer 1

Step-by-step explanation:

In a proof by contradiction we assume that it is not the case that the number 5–√ is irrational. Hence, that number is not irrational. Hence, that number is not not rational. Hence, since an even number of applications of the negation NOT to a statement does not change the statement's logical value, the working assumption is that 5–√ is a rational number, designated as x :

x=5–√(1)

Square both sides of (1):

x2=5(2)

Subtract 5 from both sides of (2):

x2−5=0(3)

to obtain an equation that has the following general form:

xn+cn−1xn−1+…+c2x2+c1x+c0=0(4)

and for which the above mentioned integral roots theorem guarantees that if (4) (or (3)) has a rational root then that rational root is an integer that divides c0 .

In our case c0=−5 .

We prove that 5–√ is not integer as follows:

4<5<9

4–√<5–√<9–√

2<5–√<3

But 2 and 3 are two consecutive natural numbers. Therefore, 5–√ is not an integer.

Since 5 is prime, the only divisors of our c0 are ±1,±5 - none of which, by manual verification, satisfy (3) and you may want to actually spell that out.

Hence, our initial assumption that 5–√ is rational was False. As such, 5–√ is irrational.

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Answer 2

Answer:

=> √7=a/b ( here a and b are co prime means they have only 1 as common factor. ... Here we find 7 is common which divide both a and b but this is contradiction because a and b are co prime they don't have common factor other than 1. So for our assumption is wrong. Hence √7 is irrational.

Also refer to the attachment.

Step-by-step explanation:

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