-0.000 000 000 742 147 676 283 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 283(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
-0.000 000 000 742 147 676 283(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:

|-0.000 000 000 742 147 676 283| = 0.000 000 000 742 147 676 283


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: 0.000 000 000 742 147 676 283.

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) 0.000 000 000 742 147 676 283 × 2 = 0 + 0.000 000 001 484 295 352 566;
  • 2) 0.000 000 001 484 295 352 566 × 2 = 0 + 0.000 000 002 968 590 705 132;
  • 3) 0.000 000 002 968 590 705 132 × 2 = 0 + 0.000 000 005 937 181 410 264;
  • 4) 0.000 000 005 937 181 410 264 × 2 = 0 + 0.000 000 011 874 362 820 528;
  • 5) 0.000 000 011 874 362 820 528 × 2 = 0 + 0.000 000 023 748 725 641 056;
  • 6) 0.000 000 023 748 725 641 056 × 2 = 0 + 0.000 000 047 497 451 282 112;
  • 7) 0.000 000 047 497 451 282 112 × 2 = 0 + 0.000 000 094 994 902 564 224;
  • 8) 0.000 000 094 994 902 564 224 × 2 = 0 + 0.000 000 189 989 805 128 448;
  • 9) 0.000 000 189 989 805 128 448 × 2 = 0 + 0.000 000 379 979 610 256 896;
  • 10) 0.000 000 379 979 610 256 896 × 2 = 0 + 0.000 000 759 959 220 513 792;
  • 11) 0.000 000 759 959 220 513 792 × 2 = 0 + 0.000 001 519 918 441 027 584;
  • 12) 0.000 001 519 918 441 027 584 × 2 = 0 + 0.000 003 039 836 882 055 168;
  • 13) 0.000 003 039 836 882 055 168 × 2 = 0 + 0.000 006 079 673 764 110 336;
  • 14) 0.000 006 079 673 764 110 336 × 2 = 0 + 0.000 012 159 347 528 220 672;
  • 15) 0.000 012 159 347 528 220 672 × 2 = 0 + 0.000 024 318 695 056 441 344;
  • 16) 0.000 024 318 695 056 441 344 × 2 = 0 + 0.000 048 637 390 112 882 688;
  • 17) 0.000 048 637 390 112 882 688 × 2 = 0 + 0.000 097 274 780 225 765 376;
  • 18) 0.000 097 274 780 225 765 376 × 2 = 0 + 0.000 194 549 560 451 530 752;
  • 19) 0.000 194 549 560 451 530 752 × 2 = 0 + 0.000 389 099 120 903 061 504;
  • 20) 0.000 389 099 120 903 061 504 × 2 = 0 + 0.000 778 198 241 806 123 008;
  • 21) 0.000 778 198 241 806 123 008 × 2 = 0 + 0.001 556 396 483 612 246 016;
  • 22) 0.001 556 396 483 612 246 016 × 2 = 0 + 0.003 112 792 967 224 492 032;
  • 23) 0.003 112 792 967 224 492 032 × 2 = 0 + 0.006 225 585 934 448 984 064;
  • 24) 0.006 225 585 934 448 984 064 × 2 = 0 + 0.012 451 171 868 897 968 128;
  • 25) 0.012 451 171 868 897 968 128 × 2 = 0 + 0.024 902 343 737 795 936 256;
  • 26) 0.024 902 343 737 795 936 256 × 2 = 0 + 0.049 804 687 475 591 872 512;
  • 27) 0.049 804 687 475 591 872 512 × 2 = 0 + 0.099 609 374 951 183 745 024;
  • 28) 0.099 609 374 951 183 745 024 × 2 = 0 + 0.199 218 749 902 367 490 048;
  • 29) 0.199 218 749 902 367 490 048 × 2 = 0 + 0.398 437 499 804 734 980 096;
  • 30) 0.398 437 499 804 734 980 096 × 2 = 0 + 0.796 874 999 609 469 960 192;
  • 31) 0.796 874 999 609 469 960 192 × 2 = 1 + 0.593 749 999 218 939 920 384;
  • 32) 0.593 749 999 218 939 920 384 × 2 = 1 + 0.187 499 998 437 879 840 768;
  • 33) 0.187 499 998 437 879 840 768 × 2 = 0 + 0.374 999 996 875 759 681 536;
  • 34) 0.374 999 996 875 759 681 536 × 2 = 0 + 0.749 999 993 751 519 363 072;
  • 35) 0.749 999 993 751 519 363 072 × 2 = 1 + 0.499 999 987 503 038 726 144;
  • 36) 0.499 999 987 503 038 726 144 × 2 = 0 + 0.999 999 975 006 077 452 288;
  • 37) 0.999 999 975 006 077 452 288 × 2 = 1 + 0.999 999 950 012 154 904 576;
  • 38) 0.999 999 950 012 154 904 576 × 2 = 1 + 0.999 999 900 024 309 809 152;
  • 39) 0.999 999 900 024 309 809 152 × 2 = 1 + 0.999 999 800 048 619 618 304;
  • 40) 0.999 999 800 048 619 618 304 × 2 = 1 + 0.999 999 600 097 239 236 608;
  • 41) 0.999 999 600 097 239 236 608 × 2 = 1 + 0.999 999 200 194 478 473 216;
  • 42) 0.999 999 200 194 478 473 216 × 2 = 1 + 0.999 998 400 388 956 946 432;
  • 43) 0.999 998 400 388 956 946 432 × 2 = 1 + 0.999 996 800 777 913 892 864;
  • 44) 0.999 996 800 777 913 892 864 × 2 = 1 + 0.999 993 601 555 827 785 728;
  • 45) 0.999 993 601 555 827 785 728 × 2 = 1 + 0.999 987 203 111 655 571 456;
  • 46) 0.999 987 203 111 655 571 456 × 2 = 1 + 0.999 974 406 223 311 142 912;
  • 47) 0.999 974 406 223 311 142 912 × 2 = 1 + 0.999 948 812 446 622 285 824;
  • 48) 0.999 948 812 446 622 285 824 × 2 = 1 + 0.999 897 624 893 244 571 648;
  • 49) 0.999 897 624 893 244 571 648 × 2 = 1 + 0.999 795 249 786 489 143 296;
  • 50) 0.999 795 249 786 489 143 296 × 2 = 1 + 0.999 590 499 572 978 286 592;
  • 51) 0.999 590 499 572 978 286 592 × 2 = 1 + 0.999 180 999 145 956 573 184;
  • 52) 0.999 180 999 145 956 573 184 × 2 = 1 + 0.998 361 998 291 913 146 368;
  • 53) 0.998 361 998 291 913 146 368 × 2 = 1 + 0.996 723 996 583 826 292 736;
  • 54) 0.996 723 996 583 826 292 736 × 2 = 1 + 0.993 447 993 167 652 585 472;

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:


0.000 000 000 742 147 676 283(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 283(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 283(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 1111 1111 111


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 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 283 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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