-0.000 000 000 742 147 676 646 671 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 671 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
-0.000 000 000 742 147 676 646 671 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:

|-0.000 000 000 742 147 676 646 671 5| = 0.000 000 000 742 147 676 646 671 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: 0.000 000 000 742 147 676 646 671 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) 0.000 000 000 742 147 676 646 671 5 × 2 = 0 + 0.000 000 001 484 295 353 293 343;
  • 2) 0.000 000 001 484 295 353 293 343 × 2 = 0 + 0.000 000 002 968 590 706 586 686;
  • 3) 0.000 000 002 968 590 706 586 686 × 2 = 0 + 0.000 000 005 937 181 413 173 372;
  • 4) 0.000 000 005 937 181 413 173 372 × 2 = 0 + 0.000 000 011 874 362 826 346 744;
  • 5) 0.000 000 011 874 362 826 346 744 × 2 = 0 + 0.000 000 023 748 725 652 693 488;
  • 6) 0.000 000 023 748 725 652 693 488 × 2 = 0 + 0.000 000 047 497 451 305 386 976;
  • 7) 0.000 000 047 497 451 305 386 976 × 2 = 0 + 0.000 000 094 994 902 610 773 952;
  • 8) 0.000 000 094 994 902 610 773 952 × 2 = 0 + 0.000 000 189 989 805 221 547 904;
  • 9) 0.000 000 189 989 805 221 547 904 × 2 = 0 + 0.000 000 379 979 610 443 095 808;
  • 10) 0.000 000 379 979 610 443 095 808 × 2 = 0 + 0.000 000 759 959 220 886 191 616;
  • 11) 0.000 000 759 959 220 886 191 616 × 2 = 0 + 0.000 001 519 918 441 772 383 232;
  • 12) 0.000 001 519 918 441 772 383 232 × 2 = 0 + 0.000 003 039 836 883 544 766 464;
  • 13) 0.000 003 039 836 883 544 766 464 × 2 = 0 + 0.000 006 079 673 767 089 532 928;
  • 14) 0.000 006 079 673 767 089 532 928 × 2 = 0 + 0.000 012 159 347 534 179 065 856;
  • 15) 0.000 012 159 347 534 179 065 856 × 2 = 0 + 0.000 024 318 695 068 358 131 712;
  • 16) 0.000 024 318 695 068 358 131 712 × 2 = 0 + 0.000 048 637 390 136 716 263 424;
  • 17) 0.000 048 637 390 136 716 263 424 × 2 = 0 + 0.000 097 274 780 273 432 526 848;
  • 18) 0.000 097 274 780 273 432 526 848 × 2 = 0 + 0.000 194 549 560 546 865 053 696;
  • 19) 0.000 194 549 560 546 865 053 696 × 2 = 0 + 0.000 389 099 121 093 730 107 392;
  • 20) 0.000 389 099 121 093 730 107 392 × 2 = 0 + 0.000 778 198 242 187 460 214 784;
  • 21) 0.000 778 198 242 187 460 214 784 × 2 = 0 + 0.001 556 396 484 374 920 429 568;
  • 22) 0.001 556 396 484 374 920 429 568 × 2 = 0 + 0.003 112 792 968 749 840 859 136;
  • 23) 0.003 112 792 968 749 840 859 136 × 2 = 0 + 0.006 225 585 937 499 681 718 272;
  • 24) 0.006 225 585 937 499 681 718 272 × 2 = 0 + 0.012 451 171 874 999 363 436 544;
  • 25) 0.012 451 171 874 999 363 436 544 × 2 = 0 + 0.024 902 343 749 998 726 873 088;
  • 26) 0.024 902 343 749 998 726 873 088 × 2 = 0 + 0.049 804 687 499 997 453 746 176;
  • 27) 0.049 804 687 499 997 453 746 176 × 2 = 0 + 0.099 609 374 999 994 907 492 352;
  • 28) 0.099 609 374 999 994 907 492 352 × 2 = 0 + 0.199 218 749 999 989 814 984 704;
  • 29) 0.199 218 749 999 989 814 984 704 × 2 = 0 + 0.398 437 499 999 979 629 969 408;
  • 30) 0.398 437 499 999 979 629 969 408 × 2 = 0 + 0.796 874 999 999 959 259 938 816;
  • 31) 0.796 874 999 999 959 259 938 816 × 2 = 1 + 0.593 749 999 999 918 519 877 632;
  • 32) 0.593 749 999 999 918 519 877 632 × 2 = 1 + 0.187 499 999 999 837 039 755 264;
  • 33) 0.187 499 999 999 837 039 755 264 × 2 = 0 + 0.374 999 999 999 674 079 510 528;
  • 34) 0.374 999 999 999 674 079 510 528 × 2 = 0 + 0.749 999 999 999 348 159 021 056;
  • 35) 0.749 999 999 999 348 159 021 056 × 2 = 1 + 0.499 999 999 998 696 318 042 112;
  • 36) 0.499 999 999 998 696 318 042 112 × 2 = 0 + 0.999 999 999 997 392 636 084 224;
  • 37) 0.999 999 999 997 392 636 084 224 × 2 = 1 + 0.999 999 999 994 785 272 168 448;
  • 38) 0.999 999 999 994 785 272 168 448 × 2 = 1 + 0.999 999 999 989 570 544 336 896;
  • 39) 0.999 999 999 989 570 544 336 896 × 2 = 1 + 0.999 999 999 979 141 088 673 792;
  • 40) 0.999 999 999 979 141 088 673 792 × 2 = 1 + 0.999 999 999 958 282 177 347 584;
  • 41) 0.999 999 999 958 282 177 347 584 × 2 = 1 + 0.999 999 999 916 564 354 695 168;
  • 42) 0.999 999 999 916 564 354 695 168 × 2 = 1 + 0.999 999 999 833 128 709 390 336;
  • 43) 0.999 999 999 833 128 709 390 336 × 2 = 1 + 0.999 999 999 666 257 418 780 672;
  • 44) 0.999 999 999 666 257 418 780 672 × 2 = 1 + 0.999 999 999 332 514 837 561 344;
  • 45) 0.999 999 999 332 514 837 561 344 × 2 = 1 + 0.999 999 998 665 029 675 122 688;
  • 46) 0.999 999 998 665 029 675 122 688 × 2 = 1 + 0.999 999 997 330 059 350 245 376;
  • 47) 0.999 999 997 330 059 350 245 376 × 2 = 1 + 0.999 999 994 660 118 700 490 752;
  • 48) 0.999 999 994 660 118 700 490 752 × 2 = 1 + 0.999 999 989 320 237 400 981 504;
  • 49) 0.999 999 989 320 237 400 981 504 × 2 = 1 + 0.999 999 978 640 474 801 963 008;
  • 50) 0.999 999 978 640 474 801 963 008 × 2 = 1 + 0.999 999 957 280 949 603 926 016;
  • 51) 0.999 999 957 280 949 603 926 016 × 2 = 1 + 0.999 999 914 561 899 207 852 032;
  • 52) 0.999 999 914 561 899 207 852 032 × 2 = 1 + 0.999 999 829 123 798 415 704 064;
  • 53) 0.999 999 829 123 798 415 704 064 × 2 = 1 + 0.999 999 658 247 596 831 408 128;
  • 54) 0.999 999 658 247 596 831 408 128 × 2 = 1 + 0.999 999 316 495 193 662 816 256;

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:


0.000 000 000 742 147 676 646 671 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 671 5(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:


0.000 000 000 742 147 676 646 671 5(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 -0.000 000 000 742 147 676 646 671 5 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