-0.000 000 000 599 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 599(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
-0.000 000 000 599(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:

|-0.000 000 000 599| = 0.000 000 000 599


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: 0.000 000 000 599.

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) 0.000 000 000 599 × 2 = 0 + 0.000 000 001 198;
  • 2) 0.000 000 001 198 × 2 = 0 + 0.000 000 002 396;
  • 3) 0.000 000 002 396 × 2 = 0 + 0.000 000 004 792;
  • 4) 0.000 000 004 792 × 2 = 0 + 0.000 000 009 584;
  • 5) 0.000 000 009 584 × 2 = 0 + 0.000 000 019 168;
  • 6) 0.000 000 019 168 × 2 = 0 + 0.000 000 038 336;
  • 7) 0.000 000 038 336 × 2 = 0 + 0.000 000 076 672;
  • 8) 0.000 000 076 672 × 2 = 0 + 0.000 000 153 344;
  • 9) 0.000 000 153 344 × 2 = 0 + 0.000 000 306 688;
  • 10) 0.000 000 306 688 × 2 = 0 + 0.000 000 613 376;
  • 11) 0.000 000 613 376 × 2 = 0 + 0.000 001 226 752;
  • 12) 0.000 001 226 752 × 2 = 0 + 0.000 002 453 504;
  • 13) 0.000 002 453 504 × 2 = 0 + 0.000 004 907 008;
  • 14) 0.000 004 907 008 × 2 = 0 + 0.000 009 814 016;
  • 15) 0.000 009 814 016 × 2 = 0 + 0.000 019 628 032;
  • 16) 0.000 019 628 032 × 2 = 0 + 0.000 039 256 064;
  • 17) 0.000 039 256 064 × 2 = 0 + 0.000 078 512 128;
  • 18) 0.000 078 512 128 × 2 = 0 + 0.000 157 024 256;
  • 19) 0.000 157 024 256 × 2 = 0 + 0.000 314 048 512;
  • 20) 0.000 314 048 512 × 2 = 0 + 0.000 628 097 024;
  • 21) 0.000 628 097 024 × 2 = 0 + 0.001 256 194 048;
  • 22) 0.001 256 194 048 × 2 = 0 + 0.002 512 388 096;
  • 23) 0.002 512 388 096 × 2 = 0 + 0.005 024 776 192;
  • 24) 0.005 024 776 192 × 2 = 0 + 0.010 049 552 384;
  • 25) 0.010 049 552 384 × 2 = 0 + 0.020 099 104 768;
  • 26) 0.020 099 104 768 × 2 = 0 + 0.040 198 209 536;
  • 27) 0.040 198 209 536 × 2 = 0 + 0.080 396 419 072;
  • 28) 0.080 396 419 072 × 2 = 0 + 0.160 792 838 144;
  • 29) 0.160 792 838 144 × 2 = 0 + 0.321 585 676 288;
  • 30) 0.321 585 676 288 × 2 = 0 + 0.643 171 352 576;
  • 31) 0.643 171 352 576 × 2 = 1 + 0.286 342 705 152;
  • 32) 0.286 342 705 152 × 2 = 0 + 0.572 685 410 304;
  • 33) 0.572 685 410 304 × 2 = 1 + 0.145 370 820 608;
  • 34) 0.145 370 820 608 × 2 = 0 + 0.290 741 641 216;
  • 35) 0.290 741 641 216 × 2 = 0 + 0.581 483 282 432;
  • 36) 0.581 483 282 432 × 2 = 1 + 0.162 966 564 864;
  • 37) 0.162 966 564 864 × 2 = 0 + 0.325 933 129 728;
  • 38) 0.325 933 129 728 × 2 = 0 + 0.651 866 259 456;
  • 39) 0.651 866 259 456 × 2 = 1 + 0.303 732 518 912;
  • 40) 0.303 732 518 912 × 2 = 0 + 0.607 465 037 824;
  • 41) 0.607 465 037 824 × 2 = 1 + 0.214 930 075 648;
  • 42) 0.214 930 075 648 × 2 = 0 + 0.429 860 151 296;
  • 43) 0.429 860 151 296 × 2 = 0 + 0.859 720 302 592;
  • 44) 0.859 720 302 592 × 2 = 1 + 0.719 440 605 184;
  • 45) 0.719 440 605 184 × 2 = 1 + 0.438 881 210 368;
  • 46) 0.438 881 210 368 × 2 = 0 + 0.877 762 420 736;
  • 47) 0.877 762 420 736 × 2 = 1 + 0.755 524 841 472;
  • 48) 0.755 524 841 472 × 2 = 1 + 0.511 049 682 944;
  • 49) 0.511 049 682 944 × 2 = 1 + 0.022 099 365 888;
  • 50) 0.022 099 365 888 × 2 = 0 + 0.044 198 731 776;
  • 51) 0.044 198 731 776 × 2 = 0 + 0.088 397 463 552;
  • 52) 0.088 397 463 552 × 2 = 0 + 0.176 794 927 104;
  • 53) 0.176 794 927 104 × 2 = 0 + 0.353 589 854 208;
  • 54) 0.353 589 854 208 × 2 = 0 + 0.707 179 708 416;

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:


0.000 000 000 599(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1001 0010 1001 1011 1000 00(2)

6. Positive number before normalization:

0.000 000 000 599(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1001 0010 1001 1011 1000 00(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:


0.000 000 000 599(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1001 0010 1001 1011 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1001 0010 1001 1011 1000 00(2) × 20 =

1.0100 1001 0100 1101 1100 000(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):
1.0100 1001 0100 1101 1100 000


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. 010 0100 1010 0110 1110 0000 =

010 0100 1010 0110 1110 0000


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) =
010 0100 1010 0110 1110 0000


Decimal number -0.000 000 000 599 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 010 0100 1010 0110 1110 0000

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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