-0.000 000 000 745 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 745 8(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
-0.000 000 000 745 8(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:

|-0.000 000 000 745 8| = 0.000 000 000 745 8


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: 0.000 000 000 745 8.

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) 0.000 000 000 745 8 × 2 = 0 + 0.000 000 001 491 6;
  • 2) 0.000 000 001 491 6 × 2 = 0 + 0.000 000 002 983 2;
  • 3) 0.000 000 002 983 2 × 2 = 0 + 0.000 000 005 966 4;
  • 4) 0.000 000 005 966 4 × 2 = 0 + 0.000 000 011 932 8;
  • 5) 0.000 000 011 932 8 × 2 = 0 + 0.000 000 023 865 6;
  • 6) 0.000 000 023 865 6 × 2 = 0 + 0.000 000 047 731 2;
  • 7) 0.000 000 047 731 2 × 2 = 0 + 0.000 000 095 462 4;
  • 8) 0.000 000 095 462 4 × 2 = 0 + 0.000 000 190 924 8;
  • 9) 0.000 000 190 924 8 × 2 = 0 + 0.000 000 381 849 6;
  • 10) 0.000 000 381 849 6 × 2 = 0 + 0.000 000 763 699 2;
  • 11) 0.000 000 763 699 2 × 2 = 0 + 0.000 001 527 398 4;
  • 12) 0.000 001 527 398 4 × 2 = 0 + 0.000 003 054 796 8;
  • 13) 0.000 003 054 796 8 × 2 = 0 + 0.000 006 109 593 6;
  • 14) 0.000 006 109 593 6 × 2 = 0 + 0.000 012 219 187 2;
  • 15) 0.000 012 219 187 2 × 2 = 0 + 0.000 024 438 374 4;
  • 16) 0.000 024 438 374 4 × 2 = 0 + 0.000 048 876 748 8;
  • 17) 0.000 048 876 748 8 × 2 = 0 + 0.000 097 753 497 6;
  • 18) 0.000 097 753 497 6 × 2 = 0 + 0.000 195 506 995 2;
  • 19) 0.000 195 506 995 2 × 2 = 0 + 0.000 391 013 990 4;
  • 20) 0.000 391 013 990 4 × 2 = 0 + 0.000 782 027 980 8;
  • 21) 0.000 782 027 980 8 × 2 = 0 + 0.001 564 055 961 6;
  • 22) 0.001 564 055 961 6 × 2 = 0 + 0.003 128 111 923 2;
  • 23) 0.003 128 111 923 2 × 2 = 0 + 0.006 256 223 846 4;
  • 24) 0.006 256 223 846 4 × 2 = 0 + 0.012 512 447 692 8;
  • 25) 0.012 512 447 692 8 × 2 = 0 + 0.025 024 895 385 6;
  • 26) 0.025 024 895 385 6 × 2 = 0 + 0.050 049 790 771 2;
  • 27) 0.050 049 790 771 2 × 2 = 0 + 0.100 099 581 542 4;
  • 28) 0.100 099 581 542 4 × 2 = 0 + 0.200 199 163 084 8;
  • 29) 0.200 199 163 084 8 × 2 = 0 + 0.400 398 326 169 6;
  • 30) 0.400 398 326 169 6 × 2 = 0 + 0.800 796 652 339 2;
  • 31) 0.800 796 652 339 2 × 2 = 1 + 0.601 593 304 678 4;
  • 32) 0.601 593 304 678 4 × 2 = 1 + 0.203 186 609 356 8;
  • 33) 0.203 186 609 356 8 × 2 = 0 + 0.406 373 218 713 6;
  • 34) 0.406 373 218 713 6 × 2 = 0 + 0.812 746 437 427 2;
  • 35) 0.812 746 437 427 2 × 2 = 1 + 0.625 492 874 854 4;
  • 36) 0.625 492 874 854 4 × 2 = 1 + 0.250 985 749 708 8;
  • 37) 0.250 985 749 708 8 × 2 = 0 + 0.501 971 499 417 6;
  • 38) 0.501 971 499 417 6 × 2 = 1 + 0.003 942 998 835 2;
  • 39) 0.003 942 998 835 2 × 2 = 0 + 0.007 885 997 670 4;
  • 40) 0.007 885 997 670 4 × 2 = 0 + 0.015 771 995 340 8;
  • 41) 0.015 771 995 340 8 × 2 = 0 + 0.031 543 990 681 6;
  • 42) 0.031 543 990 681 6 × 2 = 0 + 0.063 087 981 363 2;
  • 43) 0.063 087 981 363 2 × 2 = 0 + 0.126 175 962 726 4;
  • 44) 0.126 175 962 726 4 × 2 = 0 + 0.252 351 925 452 8;
  • 45) 0.252 351 925 452 8 × 2 = 0 + 0.504 703 850 905 6;
  • 46) 0.504 703 850 905 6 × 2 = 1 + 0.009 407 701 811 2;
  • 47) 0.009 407 701 811 2 × 2 = 0 + 0.018 815 403 622 4;
  • 48) 0.018 815 403 622 4 × 2 = 0 + 0.037 630 807 244 8;
  • 49) 0.037 630 807 244 8 × 2 = 0 + 0.075 261 614 489 6;
  • 50) 0.075 261 614 489 6 × 2 = 0 + 0.150 523 228 979 2;
  • 51) 0.150 523 228 979 2 × 2 = 0 + 0.301 046 457 958 4;
  • 52) 0.301 046 457 958 4 × 2 = 0 + 0.602 092 915 916 8;
  • 53) 0.602 092 915 916 8 × 2 = 1 + 0.204 185 831 833 6;
  • 54) 0.204 185 831 833 6 × 2 = 0 + 0.408 371 663 667 2;

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:


0.000 000 000 745 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0100 0000 0100 0000 10(2)

6. Positive number before normalization:

0.000 000 000 745 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0100 0000 0100 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:


0.000 000 000 745 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0100 0000 0100 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0100 0000 0100 0000 10(2) × 20 =

1.1001 1010 0000 0010 0000 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 1010 0000 0010 0000 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 1101 0000 0001 0000 0010 =

100 1101 0000 0001 0000 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 1101 0000 0001 0000 0010


Decimal number -0.000 000 000 745 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1101 0000 0001 0000 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