-0.000 000 000 742 128 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 128 5(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
-0.000 000 000 742 128 5(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:

|-0.000 000 000 742 128 5| = 0.000 000 000 742 128 5


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: 0.000 000 000 742 128 5.

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) 0.000 000 000 742 128 5 × 2 = 0 + 0.000 000 001 484 257;
  • 2) 0.000 000 001 484 257 × 2 = 0 + 0.000 000 002 968 514;
  • 3) 0.000 000 002 968 514 × 2 = 0 + 0.000 000 005 937 028;
  • 4) 0.000 000 005 937 028 × 2 = 0 + 0.000 000 011 874 056;
  • 5) 0.000 000 011 874 056 × 2 = 0 + 0.000 000 023 748 112;
  • 6) 0.000 000 023 748 112 × 2 = 0 + 0.000 000 047 496 224;
  • 7) 0.000 000 047 496 224 × 2 = 0 + 0.000 000 094 992 448;
  • 8) 0.000 000 094 992 448 × 2 = 0 + 0.000 000 189 984 896;
  • 9) 0.000 000 189 984 896 × 2 = 0 + 0.000 000 379 969 792;
  • 10) 0.000 000 379 969 792 × 2 = 0 + 0.000 000 759 939 584;
  • 11) 0.000 000 759 939 584 × 2 = 0 + 0.000 001 519 879 168;
  • 12) 0.000 001 519 879 168 × 2 = 0 + 0.000 003 039 758 336;
  • 13) 0.000 003 039 758 336 × 2 = 0 + 0.000 006 079 516 672;
  • 14) 0.000 006 079 516 672 × 2 = 0 + 0.000 012 159 033 344;
  • 15) 0.000 012 159 033 344 × 2 = 0 + 0.000 024 318 066 688;
  • 16) 0.000 024 318 066 688 × 2 = 0 + 0.000 048 636 133 376;
  • 17) 0.000 048 636 133 376 × 2 = 0 + 0.000 097 272 266 752;
  • 18) 0.000 097 272 266 752 × 2 = 0 + 0.000 194 544 533 504;
  • 19) 0.000 194 544 533 504 × 2 = 0 + 0.000 389 089 067 008;
  • 20) 0.000 389 089 067 008 × 2 = 0 + 0.000 778 178 134 016;
  • 21) 0.000 778 178 134 016 × 2 = 0 + 0.001 556 356 268 032;
  • 22) 0.001 556 356 268 032 × 2 = 0 + 0.003 112 712 536 064;
  • 23) 0.003 112 712 536 064 × 2 = 0 + 0.006 225 425 072 128;
  • 24) 0.006 225 425 072 128 × 2 = 0 + 0.012 450 850 144 256;
  • 25) 0.012 450 850 144 256 × 2 = 0 + 0.024 901 700 288 512;
  • 26) 0.024 901 700 288 512 × 2 = 0 + 0.049 803 400 577 024;
  • 27) 0.049 803 400 577 024 × 2 = 0 + 0.099 606 801 154 048;
  • 28) 0.099 606 801 154 048 × 2 = 0 + 0.199 213 602 308 096;
  • 29) 0.199 213 602 308 096 × 2 = 0 + 0.398 427 204 616 192;
  • 30) 0.398 427 204 616 192 × 2 = 0 + 0.796 854 409 232 384;
  • 31) 0.796 854 409 232 384 × 2 = 1 + 0.593 708 818 464 768;
  • 32) 0.593 708 818 464 768 × 2 = 1 + 0.187 417 636 929 536;
  • 33) 0.187 417 636 929 536 × 2 = 0 + 0.374 835 273 859 072;
  • 34) 0.374 835 273 859 072 × 2 = 0 + 0.749 670 547 718 144;
  • 35) 0.749 670 547 718 144 × 2 = 1 + 0.499 341 095 436 288;
  • 36) 0.499 341 095 436 288 × 2 = 0 + 0.998 682 190 872 576;
  • 37) 0.998 682 190 872 576 × 2 = 1 + 0.997 364 381 745 152;
  • 38) 0.997 364 381 745 152 × 2 = 1 + 0.994 728 763 490 304;
  • 39) 0.994 728 763 490 304 × 2 = 1 + 0.989 457 526 980 608;
  • 40) 0.989 457 526 980 608 × 2 = 1 + 0.978 915 053 961 216;
  • 41) 0.978 915 053 961 216 × 2 = 1 + 0.957 830 107 922 432;
  • 42) 0.957 830 107 922 432 × 2 = 1 + 0.915 660 215 844 864;
  • 43) 0.915 660 215 844 864 × 2 = 1 + 0.831 320 431 689 728;
  • 44) 0.831 320 431 689 728 × 2 = 1 + 0.662 640 863 379 456;
  • 45) 0.662 640 863 379 456 × 2 = 1 + 0.325 281 726 758 912;
  • 46) 0.325 281 726 758 912 × 2 = 0 + 0.650 563 453 517 824;
  • 47) 0.650 563 453 517 824 × 2 = 1 + 0.301 126 907 035 648;
  • 48) 0.301 126 907 035 648 × 2 = 0 + 0.602 253 814 071 296;
  • 49) 0.602 253 814 071 296 × 2 = 1 + 0.204 507 628 142 592;
  • 50) 0.204 507 628 142 592 × 2 = 0 + 0.409 015 256 285 184;
  • 51) 0.409 015 256 285 184 × 2 = 0 + 0.818 030 512 570 368;
  • 52) 0.818 030 512 570 368 × 2 = 1 + 0.636 061 025 140 736;
  • 53) 0.636 061 025 140 736 × 2 = 1 + 0.272 122 050 281 472;
  • 54) 0.272 122 050 281 472 × 2 = 0 + 0.544 244 100 562 944;

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:


0.000 000 000 742 128 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1001 10(2)

6. Positive number before normalization:

0.000 000 000 742 128 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1001 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:


0.000 000 000 742 128 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1001 10(2) × 20 =

1.1001 0111 1111 1101 0100 110(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 0111 1111 1101 0100 110


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 1111 1110 1010 0110 =

100 1011 1111 1110 1010 0110


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 1111 1110 1010 0110


Decimal number -0.000 000 000 742 128 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1010 0110

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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