3 years ago

Please please please urgent help❤️❤️❤️

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

5 0

Answer:

its G(2)

Step-by-step explanation:

jonny [76]3 years ago

5 0

The answer is G 2 bahaha d. Dnsisisjbd

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Oduvanchick [21]

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3 years ago

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Meter B is the answer....

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3 years ago

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Maksim231197 [3]

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3 years ago

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3 years ago

Prove that root 7 is irrational by the method of contradiction

Alchen [17]

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.

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4 years ago