-0.000 000 000 742 147 676 646 363 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 363(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
-0.000 000 000 742 147 676 646 363(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:

|-0.000 000 000 742 147 676 646 363| = 0.000 000 000 742 147 676 646 363


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: 0.000 000 000 742 147 676 646 363.

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) 0.000 000 000 742 147 676 646 363 × 2 = 0 + 0.000 000 001 484 295 353 292 726;
  • 2) 0.000 000 001 484 295 353 292 726 × 2 = 0 + 0.000 000 002 968 590 706 585 452;
  • 3) 0.000 000 002 968 590 706 585 452 × 2 = 0 + 0.000 000 005 937 181 413 170 904;
  • 4) 0.000 000 005 937 181 413 170 904 × 2 = 0 + 0.000 000 011 874 362 826 341 808;
  • 5) 0.000 000 011 874 362 826 341 808 × 2 = 0 + 0.000 000 023 748 725 652 683 616;
  • 6) 0.000 000 023 748 725 652 683 616 × 2 = 0 + 0.000 000 047 497 451 305 367 232;
  • 7) 0.000 000 047 497 451 305 367 232 × 2 = 0 + 0.000 000 094 994 902 610 734 464;
  • 8) 0.000 000 094 994 902 610 734 464 × 2 = 0 + 0.000 000 189 989 805 221 468 928;
  • 9) 0.000 000 189 989 805 221 468 928 × 2 = 0 + 0.000 000 379 979 610 442 937 856;
  • 10) 0.000 000 379 979 610 442 937 856 × 2 = 0 + 0.000 000 759 959 220 885 875 712;
  • 11) 0.000 000 759 959 220 885 875 712 × 2 = 0 + 0.000 001 519 918 441 771 751 424;
  • 12) 0.000 001 519 918 441 771 751 424 × 2 = 0 + 0.000 003 039 836 883 543 502 848;
  • 13) 0.000 003 039 836 883 543 502 848 × 2 = 0 + 0.000 006 079 673 767 087 005 696;
  • 14) 0.000 006 079 673 767 087 005 696 × 2 = 0 + 0.000 012 159 347 534 174 011 392;
  • 15) 0.000 012 159 347 534 174 011 392 × 2 = 0 + 0.000 024 318 695 068 348 022 784;
  • 16) 0.000 024 318 695 068 348 022 784 × 2 = 0 + 0.000 048 637 390 136 696 045 568;
  • 17) 0.000 048 637 390 136 696 045 568 × 2 = 0 + 0.000 097 274 780 273 392 091 136;
  • 18) 0.000 097 274 780 273 392 091 136 × 2 = 0 + 0.000 194 549 560 546 784 182 272;
  • 19) 0.000 194 549 560 546 784 182 272 × 2 = 0 + 0.000 389 099 121 093 568 364 544;
  • 20) 0.000 389 099 121 093 568 364 544 × 2 = 0 + 0.000 778 198 242 187 136 729 088;
  • 21) 0.000 778 198 242 187 136 729 088 × 2 = 0 + 0.001 556 396 484 374 273 458 176;
  • 22) 0.001 556 396 484 374 273 458 176 × 2 = 0 + 0.003 112 792 968 748 546 916 352;
  • 23) 0.003 112 792 968 748 546 916 352 × 2 = 0 + 0.006 225 585 937 497 093 832 704;
  • 24) 0.006 225 585 937 497 093 832 704 × 2 = 0 + 0.012 451 171 874 994 187 665 408;
  • 25) 0.012 451 171 874 994 187 665 408 × 2 = 0 + 0.024 902 343 749 988 375 330 816;
  • 26) 0.024 902 343 749 988 375 330 816 × 2 = 0 + 0.049 804 687 499 976 750 661 632;
  • 27) 0.049 804 687 499 976 750 661 632 × 2 = 0 + 0.099 609 374 999 953 501 323 264;
  • 28) 0.099 609 374 999 953 501 323 264 × 2 = 0 + 0.199 218 749 999 907 002 646 528;
  • 29) 0.199 218 749 999 907 002 646 528 × 2 = 0 + 0.398 437 499 999 814 005 293 056;
  • 30) 0.398 437 499 999 814 005 293 056 × 2 = 0 + 0.796 874 999 999 628 010 586 112;
  • 31) 0.796 874 999 999 628 010 586 112 × 2 = 1 + 0.593 749 999 999 256 021 172 224;
  • 32) 0.593 749 999 999 256 021 172 224 × 2 = 1 + 0.187 499 999 998 512 042 344 448;
  • 33) 0.187 499 999 998 512 042 344 448 × 2 = 0 + 0.374 999 999 997 024 084 688 896;
  • 34) 0.374 999 999 997 024 084 688 896 × 2 = 0 + 0.749 999 999 994 048 169 377 792;
  • 35) 0.749 999 999 994 048 169 377 792 × 2 = 1 + 0.499 999 999 988 096 338 755 584;
  • 36) 0.499 999 999 988 096 338 755 584 × 2 = 0 + 0.999 999 999 976 192 677 511 168;
  • 37) 0.999 999 999 976 192 677 511 168 × 2 = 1 + 0.999 999 999 952 385 355 022 336;
  • 38) 0.999 999 999 952 385 355 022 336 × 2 = 1 + 0.999 999 999 904 770 710 044 672;
  • 39) 0.999 999 999 904 770 710 044 672 × 2 = 1 + 0.999 999 999 809 541 420 089 344;
  • 40) 0.999 999 999 809 541 420 089 344 × 2 = 1 + 0.999 999 999 619 082 840 178 688;
  • 41) 0.999 999 999 619 082 840 178 688 × 2 = 1 + 0.999 999 999 238 165 680 357 376;
  • 42) 0.999 999 999 238 165 680 357 376 × 2 = 1 + 0.999 999 998 476 331 360 714 752;
  • 43) 0.999 999 998 476 331 360 714 752 × 2 = 1 + 0.999 999 996 952 662 721 429 504;
  • 44) 0.999 999 996 952 662 721 429 504 × 2 = 1 + 0.999 999 993 905 325 442 859 008;
  • 45) 0.999 999 993 905 325 442 859 008 × 2 = 1 + 0.999 999 987 810 650 885 718 016;
  • 46) 0.999 999 987 810 650 885 718 016 × 2 = 1 + 0.999 999 975 621 301 771 436 032;
  • 47) 0.999 999 975 621 301 771 436 032 × 2 = 1 + 0.999 999 951 242 603 542 872 064;
  • 48) 0.999 999 951 242 603 542 872 064 × 2 = 1 + 0.999 999 902 485 207 085 744 128;
  • 49) 0.999 999 902 485 207 085 744 128 × 2 = 1 + 0.999 999 804 970 414 171 488 256;
  • 50) 0.999 999 804 970 414 171 488 256 × 2 = 1 + 0.999 999 609 940 828 342 976 512;
  • 51) 0.999 999 609 940 828 342 976 512 × 2 = 1 + 0.999 999 219 881 656 685 953 024;
  • 52) 0.999 999 219 881 656 685 953 024 × 2 = 1 + 0.999 998 439 763 313 371 906 048;
  • 53) 0.999 998 439 763 313 371 906 048 × 2 = 1 + 0.999 996 879 526 626 743 812 096;
  • 54) 0.999 996 879 526 626 743 812 096 × 2 = 1 + 0.999 993 759 053 253 487 624 192;

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:


0.000 000 000 742 147 676 646 363(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 363(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:


0.000 000 000 742 147 676 646 363(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 -0.000 000 000 742 147 676 646 363 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