-0.000 000 110 943 801 701 068 878 226 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 110 943 801 701 068 878 226(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
-0.000 000 110 943 801 701 068 878 226(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:

|-0.000 000 110 943 801 701 068 878 226| = 0.000 000 110 943 801 701 068 878 226


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: 0.000 000 110 943 801 701 068 878 226.

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) 0.000 000 110 943 801 701 068 878 226 × 2 = 0 + 0.000 000 221 887 603 402 137 756 452;
  • 2) 0.000 000 221 887 603 402 137 756 452 × 2 = 0 + 0.000 000 443 775 206 804 275 512 904;
  • 3) 0.000 000 443 775 206 804 275 512 904 × 2 = 0 + 0.000 000 887 550 413 608 551 025 808;
  • 4) 0.000 000 887 550 413 608 551 025 808 × 2 = 0 + 0.000 001 775 100 827 217 102 051 616;
  • 5) 0.000 001 775 100 827 217 102 051 616 × 2 = 0 + 0.000 003 550 201 654 434 204 103 232;
  • 6) 0.000 003 550 201 654 434 204 103 232 × 2 = 0 + 0.000 007 100 403 308 868 408 206 464;
  • 7) 0.000 007 100 403 308 868 408 206 464 × 2 = 0 + 0.000 014 200 806 617 736 816 412 928;
  • 8) 0.000 014 200 806 617 736 816 412 928 × 2 = 0 + 0.000 028 401 613 235 473 632 825 856;
  • 9) 0.000 028 401 613 235 473 632 825 856 × 2 = 0 + 0.000 056 803 226 470 947 265 651 712;
  • 10) 0.000 056 803 226 470 947 265 651 712 × 2 = 0 + 0.000 113 606 452 941 894 531 303 424;
  • 11) 0.000 113 606 452 941 894 531 303 424 × 2 = 0 + 0.000 227 212 905 883 789 062 606 848;
  • 12) 0.000 227 212 905 883 789 062 606 848 × 2 = 0 + 0.000 454 425 811 767 578 125 213 696;
  • 13) 0.000 454 425 811 767 578 125 213 696 × 2 = 0 + 0.000 908 851 623 535 156 250 427 392;
  • 14) 0.000 908 851 623 535 156 250 427 392 × 2 = 0 + 0.001 817 703 247 070 312 500 854 784;
  • 15) 0.001 817 703 247 070 312 500 854 784 × 2 = 0 + 0.003 635 406 494 140 625 001 709 568;
  • 16) 0.003 635 406 494 140 625 001 709 568 × 2 = 0 + 0.007 270 812 988 281 250 003 419 136;
  • 17) 0.007 270 812 988 281 250 003 419 136 × 2 = 0 + 0.014 541 625 976 562 500 006 838 272;
  • 18) 0.014 541 625 976 562 500 006 838 272 × 2 = 0 + 0.029 083 251 953 125 000 013 676 544;
  • 19) 0.029 083 251 953 125 000 013 676 544 × 2 = 0 + 0.058 166 503 906 250 000 027 353 088;
  • 20) 0.058 166 503 906 250 000 027 353 088 × 2 = 0 + 0.116 333 007 812 500 000 054 706 176;
  • 21) 0.116 333 007 812 500 000 054 706 176 × 2 = 0 + 0.232 666 015 625 000 000 109 412 352;
  • 22) 0.232 666 015 625 000 000 109 412 352 × 2 = 0 + 0.465 332 031 250 000 000 218 824 704;
  • 23) 0.465 332 031 250 000 000 218 824 704 × 2 = 0 + 0.930 664 062 500 000 000 437 649 408;
  • 24) 0.930 664 062 500 000 000 437 649 408 × 2 = 1 + 0.861 328 125 000 000 000 875 298 816;
  • 25) 0.861 328 125 000 000 000 875 298 816 × 2 = 1 + 0.722 656 250 000 000 001 750 597 632;
  • 26) 0.722 656 250 000 000 001 750 597 632 × 2 = 1 + 0.445 312 500 000 000 003 501 195 264;
  • 27) 0.445 312 500 000 000 003 501 195 264 × 2 = 0 + 0.890 625 000 000 000 007 002 390 528;
  • 28) 0.890 625 000 000 000 007 002 390 528 × 2 = 1 + 0.781 250 000 000 000 014 004 781 056;
  • 29) 0.781 250 000 000 000 014 004 781 056 × 2 = 1 + 0.562 500 000 000 000 028 009 562 112;
  • 30) 0.562 500 000 000 000 028 009 562 112 × 2 = 1 + 0.125 000 000 000 000 056 019 124 224;
  • 31) 0.125 000 000 000 000 056 019 124 224 × 2 = 0 + 0.250 000 000 000 000 112 038 248 448;
  • 32) 0.250 000 000 000 000 112 038 248 448 × 2 = 0 + 0.500 000 000 000 000 224 076 496 896;
  • 33) 0.500 000 000 000 000 224 076 496 896 × 2 = 1 + 0.000 000 000 000 000 448 152 993 792;
  • 34) 0.000 000 000 000 000 448 152 993 792 × 2 = 0 + 0.000 000 000 000 000 896 305 987 584;
  • 35) 0.000 000 000 000 000 896 305 987 584 × 2 = 0 + 0.000 000 000 000 001 792 611 975 168;
  • 36) 0.000 000 000 000 001 792 611 975 168 × 2 = 0 + 0.000 000 000 000 003 585 223 950 336;
  • 37) 0.000 000 000 000 003 585 223 950 336 × 2 = 0 + 0.000 000 000 000 007 170 447 900 672;
  • 38) 0.000 000 000 000 007 170 447 900 672 × 2 = 0 + 0.000 000 000 000 014 340 895 801 344;
  • 39) 0.000 000 000 000 014 340 895 801 344 × 2 = 0 + 0.000 000 000 000 028 681 791 602 688;
  • 40) 0.000 000 000 000 028 681 791 602 688 × 2 = 0 + 0.000 000 000 000 057 363 583 205 376;
  • 41) 0.000 000 000 000 057 363 583 205 376 × 2 = 0 + 0.000 000 000 000 114 727 166 410 752;
  • 42) 0.000 000 000 000 114 727 166 410 752 × 2 = 0 + 0.000 000 000 000 229 454 332 821 504;
  • 43) 0.000 000 000 000 229 454 332 821 504 × 2 = 0 + 0.000 000 000 000 458 908 665 643 008;
  • 44) 0.000 000 000 000 458 908 665 643 008 × 2 = 0 + 0.000 000 000 000 917 817 331 286 016;
  • 45) 0.000 000 000 000 917 817 331 286 016 × 2 = 0 + 0.000 000 000 001 835 634 662 572 032;
  • 46) 0.000 000 000 001 835 634 662 572 032 × 2 = 0 + 0.000 000 000 003 671 269 325 144 064;
  • 47) 0.000 000 000 003 671 269 325 144 064 × 2 = 0 + 0.000 000 000 007 342 538 650 288 128;

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:


0.000 000 110 943 801 701 068 878 226(10) =

0.0000 0000 0000 0000 0000 0001 1101 1100 1000 0000 0000 000(2)

6. Positive number before normalization:

0.000 000 110 943 801 701 068 878 226(10) =

0.0000 0000 0000 0000 0000 0001 1101 1100 1000 0000 0000 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 943 801 701 068 878 226(10) =

0.0000 0000 0000 0000 0000 0001 1101 1100 1000 0000 0000 000(2) =

0.0000 0000 0000 0000 0000 0001 1101 1100 1000 0000 0000 000(2) × 20 =

1.1101 1100 1000 0000 0000 000(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1100 1000 0000 0000 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 110 1110 0100 0000 0000 0000 =

110 1110 0100 0000 0000 0000


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 0111

Mantissa (23 bits) =
110 1110 0100 0000 0000 0000


Decimal number -0.000 000 110 943 801 701 068 878 226 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 110 1110 0100 0000 0000 0000

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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