Let assume that $\sqrt7$ is a rational number. Therefore it can be expressed as a fraction $\dfrac{a}{b}$ where $a,b\in\mathbb{Z}$ and $\text{gcd}(a,b)=1$ .

$\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2$

This means that $a^2$ is divisible by 7, and therefore also $a$ is divisible by 7.

So, $a=7k$ where $k\in\mathbb{Z}$

$(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2$

Analogically to $a^2=7b^2$ ------- $b^2$ is divisible by 7 and therefore so is $b$ .

But if both numbers $a$ and $b$ are divisible by 7, then $\text{gcd}(a,b)=7$ which contradicts our earlier assumption that $\text{gcd}(a,b)=1$ .

Therefore $\sqrt7$ is an irrational number.

8 0

4 years ago

Shayla wants to refinish the floors in her kitchen. Shayla predicts she will be able to cover the room with 75 square feet of ti

iren [92.7K]

Answer:

Yes, the area of the floor space in the room is less than 54 square feet.

Step-by-step explanation:

here is a picture of it on quizzez

3 0

3 years ago

Solve for x: 4 = (-8) +3x. Will you please help?

Zanzabum

8 + 4 = 3x
12 = 3x
divide both sides by 3
x= 4

8 0

3 years ago

Read 2 more answers