-0.000 000 000 739 97 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 739 97(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 739 97(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 739 97| = 0.000 000 000 739 97


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 739 97.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 739 97 × 2 = 0 + 0.000 000 001 479 94;
  • 2) 0.000 000 001 479 94 × 2 = 0 + 0.000 000 002 959 88;
  • 3) 0.000 000 002 959 88 × 2 = 0 + 0.000 000 005 919 76;
  • 4) 0.000 000 005 919 76 × 2 = 0 + 0.000 000 011 839 52;
  • 5) 0.000 000 011 839 52 × 2 = 0 + 0.000 000 023 679 04;
  • 6) 0.000 000 023 679 04 × 2 = 0 + 0.000 000 047 358 08;
  • 7) 0.000 000 047 358 08 × 2 = 0 + 0.000 000 094 716 16;
  • 8) 0.000 000 094 716 16 × 2 = 0 + 0.000 000 189 432 32;
  • 9) 0.000 000 189 432 32 × 2 = 0 + 0.000 000 378 864 64;
  • 10) 0.000 000 378 864 64 × 2 = 0 + 0.000 000 757 729 28;
  • 11) 0.000 000 757 729 28 × 2 = 0 + 0.000 001 515 458 56;
  • 12) 0.000 001 515 458 56 × 2 = 0 + 0.000 003 030 917 12;
  • 13) 0.000 003 030 917 12 × 2 = 0 + 0.000 006 061 834 24;
  • 14) 0.000 006 061 834 24 × 2 = 0 + 0.000 012 123 668 48;
  • 15) 0.000 012 123 668 48 × 2 = 0 + 0.000 024 247 336 96;
  • 16) 0.000 024 247 336 96 × 2 = 0 + 0.000 048 494 673 92;
  • 17) 0.000 048 494 673 92 × 2 = 0 + 0.000 096 989 347 84;
  • 18) 0.000 096 989 347 84 × 2 = 0 + 0.000 193 978 695 68;
  • 19) 0.000 193 978 695 68 × 2 = 0 + 0.000 387 957 391 36;
  • 20) 0.000 387 957 391 36 × 2 = 0 + 0.000 775 914 782 72;
  • 21) 0.000 775 914 782 72 × 2 = 0 + 0.001 551 829 565 44;
  • 22) 0.001 551 829 565 44 × 2 = 0 + 0.003 103 659 130 88;
  • 23) 0.003 103 659 130 88 × 2 = 0 + 0.006 207 318 261 76;
  • 24) 0.006 207 318 261 76 × 2 = 0 + 0.012 414 636 523 52;
  • 25) 0.012 414 636 523 52 × 2 = 0 + 0.024 829 273 047 04;
  • 26) 0.024 829 273 047 04 × 2 = 0 + 0.049 658 546 094 08;
  • 27) 0.049 658 546 094 08 × 2 = 0 + 0.099 317 092 188 16;
  • 28) 0.099 317 092 188 16 × 2 = 0 + 0.198 634 184 376 32;
  • 29) 0.198 634 184 376 32 × 2 = 0 + 0.397 268 368 752 64;
  • 30) 0.397 268 368 752 64 × 2 = 0 + 0.794 536 737 505 28;
  • 31) 0.794 536 737 505 28 × 2 = 1 + 0.589 073 475 010 56;
  • 32) 0.589 073 475 010 56 × 2 = 1 + 0.178 146 950 021 12;
  • 33) 0.178 146 950 021 12 × 2 = 0 + 0.356 293 900 042 24;
  • 34) 0.356 293 900 042 24 × 2 = 0 + 0.712 587 800 084 48;
  • 35) 0.712 587 800 084 48 × 2 = 1 + 0.425 175 600 168 96;
  • 36) 0.425 175 600 168 96 × 2 = 0 + 0.850 351 200 337 92;
  • 37) 0.850 351 200 337 92 × 2 = 1 + 0.700 702 400 675 84;
  • 38) 0.700 702 400 675 84 × 2 = 1 + 0.401 404 801 351 68;
  • 39) 0.401 404 801 351 68 × 2 = 0 + 0.802 809 602 703 36;
  • 40) 0.802 809 602 703 36 × 2 = 1 + 0.605 619 205 406 72;
  • 41) 0.605 619 205 406 72 × 2 = 1 + 0.211 238 410 813 44;
  • 42) 0.211 238 410 813 44 × 2 = 0 + 0.422 476 821 626 88;
  • 43) 0.422 476 821 626 88 × 2 = 0 + 0.844 953 643 253 76;
  • 44) 0.844 953 643 253 76 × 2 = 1 + 0.689 907 286 507 52;
  • 45) 0.689 907 286 507 52 × 2 = 1 + 0.379 814 573 015 04;
  • 46) 0.379 814 573 015 04 × 2 = 0 + 0.759 629 146 030 08;
  • 47) 0.759 629 146 030 08 × 2 = 1 + 0.519 258 292 060 16;
  • 48) 0.519 258 292 060 16 × 2 = 1 + 0.038 516 584 120 32;
  • 49) 0.038 516 584 120 32 × 2 = 0 + 0.077 033 168 240 64;
  • 50) 0.077 033 168 240 64 × 2 = 0 + 0.154 066 336 481 28;
  • 51) 0.154 066 336 481 28 × 2 = 0 + 0.308 132 672 962 56;
  • 52) 0.308 132 672 962 56 × 2 = 0 + 0.616 265 345 925 12;
  • 53) 0.616 265 345 925 12 × 2 = 1 + 0.232 530 691 850 24;
  • 54) 0.232 530 691 850 24 × 2 = 0 + 0.465 061 383 700 48;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 739 97(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1001 1011 0000 10(2)

6. Positive number before normalization:

0.000 000 000 739 97(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1001 1011 0000 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 739 97(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1001 1011 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1001 1011 0000 10(2) × 20 =

1.1001 0110 1100 1101 1000 010(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0110 1100 1101 1000 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 0110 0110 1100 0010 =

100 1011 0110 0110 1100 0010


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 0110 0110 1100 0010


Decimal number -0.000 000 000 739 97 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 0110 0110 1100 0010

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111