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

Convert decimal -0.000 000 000 742 147 676 646 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 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 5| = 0.000 000 000 742 147 676 646 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 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 5 × 2 = 0 + 0.000 000 001 484 295 353 293;
  • 2) 0.000 000 001 484 295 353 293 × 2 = 0 + 0.000 000 002 968 590 706 586;
  • 3) 0.000 000 002 968 590 706 586 × 2 = 0 + 0.000 000 005 937 181 413 172;
  • 4) 0.000 000 005 937 181 413 172 × 2 = 0 + 0.000 000 011 874 362 826 344;
  • 5) 0.000 000 011 874 362 826 344 × 2 = 0 + 0.000 000 023 748 725 652 688;
  • 6) 0.000 000 023 748 725 652 688 × 2 = 0 + 0.000 000 047 497 451 305 376;
  • 7) 0.000 000 047 497 451 305 376 × 2 = 0 + 0.000 000 094 994 902 610 752;
  • 8) 0.000 000 094 994 902 610 752 × 2 = 0 + 0.000 000 189 989 805 221 504;
  • 9) 0.000 000 189 989 805 221 504 × 2 = 0 + 0.000 000 379 979 610 443 008;
  • 10) 0.000 000 379 979 610 443 008 × 2 = 0 + 0.000 000 759 959 220 886 016;
  • 11) 0.000 000 759 959 220 886 016 × 2 = 0 + 0.000 001 519 918 441 772 032;
  • 12) 0.000 001 519 918 441 772 032 × 2 = 0 + 0.000 003 039 836 883 544 064;
  • 13) 0.000 003 039 836 883 544 064 × 2 = 0 + 0.000 006 079 673 767 088 128;
  • 14) 0.000 006 079 673 767 088 128 × 2 = 0 + 0.000 012 159 347 534 176 256;
  • 15) 0.000 012 159 347 534 176 256 × 2 = 0 + 0.000 024 318 695 068 352 512;
  • 16) 0.000 024 318 695 068 352 512 × 2 = 0 + 0.000 048 637 390 136 705 024;
  • 17) 0.000 048 637 390 136 705 024 × 2 = 0 + 0.000 097 274 780 273 410 048;
  • 18) 0.000 097 274 780 273 410 048 × 2 = 0 + 0.000 194 549 560 546 820 096;
  • 19) 0.000 194 549 560 546 820 096 × 2 = 0 + 0.000 389 099 121 093 640 192;
  • 20) 0.000 389 099 121 093 640 192 × 2 = 0 + 0.000 778 198 242 187 280 384;
  • 21) 0.000 778 198 242 187 280 384 × 2 = 0 + 0.001 556 396 484 374 560 768;
  • 22) 0.001 556 396 484 374 560 768 × 2 = 0 + 0.003 112 792 968 749 121 536;
  • 23) 0.003 112 792 968 749 121 536 × 2 = 0 + 0.006 225 585 937 498 243 072;
  • 24) 0.006 225 585 937 498 243 072 × 2 = 0 + 0.012 451 171 874 996 486 144;
  • 25) 0.012 451 171 874 996 486 144 × 2 = 0 + 0.024 902 343 749 992 972 288;
  • 26) 0.024 902 343 749 992 972 288 × 2 = 0 + 0.049 804 687 499 985 944 576;
  • 27) 0.049 804 687 499 985 944 576 × 2 = 0 + 0.099 609 374 999 971 889 152;
  • 28) 0.099 609 374 999 971 889 152 × 2 = 0 + 0.199 218 749 999 943 778 304;
  • 29) 0.199 218 749 999 943 778 304 × 2 = 0 + 0.398 437 499 999 887 556 608;
  • 30) 0.398 437 499 999 887 556 608 × 2 = 0 + 0.796 874 999 999 775 113 216;
  • 31) 0.796 874 999 999 775 113 216 × 2 = 1 + 0.593 749 999 999 550 226 432;
  • 32) 0.593 749 999 999 550 226 432 × 2 = 1 + 0.187 499 999 999 100 452 864;
  • 33) 0.187 499 999 999 100 452 864 × 2 = 0 + 0.374 999 999 998 200 905 728;
  • 34) 0.374 999 999 998 200 905 728 × 2 = 0 + 0.749 999 999 996 401 811 456;
  • 35) 0.749 999 999 996 401 811 456 × 2 = 1 + 0.499 999 999 992 803 622 912;
  • 36) 0.499 999 999 992 803 622 912 × 2 = 0 + 0.999 999 999 985 607 245 824;
  • 37) 0.999 999 999 985 607 245 824 × 2 = 1 + 0.999 999 999 971 214 491 648;
  • 38) 0.999 999 999 971 214 491 648 × 2 = 1 + 0.999 999 999 942 428 983 296;
  • 39) 0.999 999 999 942 428 983 296 × 2 = 1 + 0.999 999 999 884 857 966 592;
  • 40) 0.999 999 999 884 857 966 592 × 2 = 1 + 0.999 999 999 769 715 933 184;
  • 41) 0.999 999 999 769 715 933 184 × 2 = 1 + 0.999 999 999 539 431 866 368;
  • 42) 0.999 999 999 539 431 866 368 × 2 = 1 + 0.999 999 999 078 863 732 736;
  • 43) 0.999 999 999 078 863 732 736 × 2 = 1 + 0.999 999 998 157 727 465 472;
  • 44) 0.999 999 998 157 727 465 472 × 2 = 1 + 0.999 999 996 315 454 930 944;
  • 45) 0.999 999 996 315 454 930 944 × 2 = 1 + 0.999 999 992 630 909 861 888;
  • 46) 0.999 999 992 630 909 861 888 × 2 = 1 + 0.999 999 985 261 819 723 776;
  • 47) 0.999 999 985 261 819 723 776 × 2 = 1 + 0.999 999 970 523 639 447 552;
  • 48) 0.999 999 970 523 639 447 552 × 2 = 1 + 0.999 999 941 047 278 895 104;
  • 49) 0.999 999 941 047 278 895 104 × 2 = 1 + 0.999 999 882 094 557 790 208;
  • 50) 0.999 999 882 094 557 790 208 × 2 = 1 + 0.999 999 764 189 115 580 416;
  • 51) 0.999 999 764 189 115 580 416 × 2 = 1 + 0.999 999 528 378 231 160 832;
  • 52) 0.999 999 528 378 231 160 832 × 2 = 1 + 0.999 999 056 756 462 321 664;
  • 53) 0.999 999 056 756 462 321 664 × 2 = 1 + 0.999 998 113 512 924 643 328;
  • 54) 0.999 998 113 512 924 643 328 × 2 = 1 + 0.999 996 227 025 849 286 656;

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 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 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 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 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