0.000 000 060 536 876 844 707 876 443 862 825 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 060 536 876 844 707 876 443 862 825(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
0.000 000 060 536 876 844 707 876 443 862 825(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 060 536 876 844 707 876 443 862 825.

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) 0.000 000 060 536 876 844 707 876 443 862 825 × 2 = 0 + 0.000 000 121 073 753 689 415 752 887 725 65;
  • 2) 0.000 000 121 073 753 689 415 752 887 725 65 × 2 = 0 + 0.000 000 242 147 507 378 831 505 775 451 3;
  • 3) 0.000 000 242 147 507 378 831 505 775 451 3 × 2 = 0 + 0.000 000 484 295 014 757 663 011 550 902 6;
  • 4) 0.000 000 484 295 014 757 663 011 550 902 6 × 2 = 0 + 0.000 000 968 590 029 515 326 023 101 805 2;
  • 5) 0.000 000 968 590 029 515 326 023 101 805 2 × 2 = 0 + 0.000 001 937 180 059 030 652 046 203 610 4;
  • 6) 0.000 001 937 180 059 030 652 046 203 610 4 × 2 = 0 + 0.000 003 874 360 118 061 304 092 407 220 8;
  • 7) 0.000 003 874 360 118 061 304 092 407 220 8 × 2 = 0 + 0.000 007 748 720 236 122 608 184 814 441 6;
  • 8) 0.000 007 748 720 236 122 608 184 814 441 6 × 2 = 0 + 0.000 015 497 440 472 245 216 369 628 883 2;
  • 9) 0.000 015 497 440 472 245 216 369 628 883 2 × 2 = 0 + 0.000 030 994 880 944 490 432 739 257 766 4;
  • 10) 0.000 030 994 880 944 490 432 739 257 766 4 × 2 = 0 + 0.000 061 989 761 888 980 865 478 515 532 8;
  • 11) 0.000 061 989 761 888 980 865 478 515 532 8 × 2 = 0 + 0.000 123 979 523 777 961 730 957 031 065 6;
  • 12) 0.000 123 979 523 777 961 730 957 031 065 6 × 2 = 0 + 0.000 247 959 047 555 923 461 914 062 131 2;
  • 13) 0.000 247 959 047 555 923 461 914 062 131 2 × 2 = 0 + 0.000 495 918 095 111 846 923 828 124 262 4;
  • 14) 0.000 495 918 095 111 846 923 828 124 262 4 × 2 = 0 + 0.000 991 836 190 223 693 847 656 248 524 8;
  • 15) 0.000 991 836 190 223 693 847 656 248 524 8 × 2 = 0 + 0.001 983 672 380 447 387 695 312 497 049 6;
  • 16) 0.001 983 672 380 447 387 695 312 497 049 6 × 2 = 0 + 0.003 967 344 760 894 775 390 624 994 099 2;
  • 17) 0.003 967 344 760 894 775 390 624 994 099 2 × 2 = 0 + 0.007 934 689 521 789 550 781 249 988 198 4;
  • 18) 0.007 934 689 521 789 550 781 249 988 198 4 × 2 = 0 + 0.015 869 379 043 579 101 562 499 976 396 8;
  • 19) 0.015 869 379 043 579 101 562 499 976 396 8 × 2 = 0 + 0.031 738 758 087 158 203 124 999 952 793 6;
  • 20) 0.031 738 758 087 158 203 124 999 952 793 6 × 2 = 0 + 0.063 477 516 174 316 406 249 999 905 587 2;
  • 21) 0.063 477 516 174 316 406 249 999 905 587 2 × 2 = 0 + 0.126 955 032 348 632 812 499 999 811 174 4;
  • 22) 0.126 955 032 348 632 812 499 999 811 174 4 × 2 = 0 + 0.253 910 064 697 265 624 999 999 622 348 8;
  • 23) 0.253 910 064 697 265 624 999 999 622 348 8 × 2 = 0 + 0.507 820 129 394 531 249 999 999 244 697 6;
  • 24) 0.507 820 129 394 531 249 999 999 244 697 6 × 2 = 1 + 0.015 640 258 789 062 499 999 998 489 395 2;
  • 25) 0.015 640 258 789 062 499 999 998 489 395 2 × 2 = 0 + 0.031 280 517 578 124 999 999 996 978 790 4;
  • 26) 0.031 280 517 578 124 999 999 996 978 790 4 × 2 = 0 + 0.062 561 035 156 249 999 999 993 957 580 8;
  • 27) 0.062 561 035 156 249 999 999 993 957 580 8 × 2 = 0 + 0.125 122 070 312 499 999 999 987 915 161 6;
  • 28) 0.125 122 070 312 499 999 999 987 915 161 6 × 2 = 0 + 0.250 244 140 624 999 999 999 975 830 323 2;
  • 29) 0.250 244 140 624 999 999 999 975 830 323 2 × 2 = 0 + 0.500 488 281 249 999 999 999 951 660 646 4;
  • 30) 0.500 488 281 249 999 999 999 951 660 646 4 × 2 = 1 + 0.000 976 562 499 999 999 999 903 321 292 8;
  • 31) 0.000 976 562 499 999 999 999 903 321 292 8 × 2 = 0 + 0.001 953 124 999 999 999 999 806 642 585 6;
  • 32) 0.001 953 124 999 999 999 999 806 642 585 6 × 2 = 0 + 0.003 906 249 999 999 999 999 613 285 171 2;
  • 33) 0.003 906 249 999 999 999 999 613 285 171 2 × 2 = 0 + 0.007 812 499 999 999 999 999 226 570 342 4;
  • 34) 0.007 812 499 999 999 999 999 226 570 342 4 × 2 = 0 + 0.015 624 999 999 999 999 998 453 140 684 8;
  • 35) 0.015 624 999 999 999 999 998 453 140 684 8 × 2 = 0 + 0.031 249 999 999 999 999 996 906 281 369 6;
  • 36) 0.031 249 999 999 999 999 996 906 281 369 6 × 2 = 0 + 0.062 499 999 999 999 999 993 812 562 739 2;
  • 37) 0.062 499 999 999 999 999 993 812 562 739 2 × 2 = 0 + 0.124 999 999 999 999 999 987 625 125 478 4;
  • 38) 0.124 999 999 999 999 999 987 625 125 478 4 × 2 = 0 + 0.249 999 999 999 999 999 975 250 250 956 8;
  • 39) 0.249 999 999 999 999 999 975 250 250 956 8 × 2 = 0 + 0.499 999 999 999 999 999 950 500 501 913 6;
  • 40) 0.499 999 999 999 999 999 950 500 501 913 6 × 2 = 0 + 0.999 999 999 999 999 999 901 001 003 827 2;
  • 41) 0.999 999 999 999 999 999 901 001 003 827 2 × 2 = 1 + 0.999 999 999 999 999 999 802 002 007 654 4;
  • 42) 0.999 999 999 999 999 999 802 002 007 654 4 × 2 = 1 + 0.999 999 999 999 999 999 604 004 015 308 8;
  • 43) 0.999 999 999 999 999 999 604 004 015 308 8 × 2 = 1 + 0.999 999 999 999 999 999 208 008 030 617 6;
  • 44) 0.999 999 999 999 999 999 208 008 030 617 6 × 2 = 1 + 0.999 999 999 999 999 998 416 016 061 235 2;
  • 45) 0.999 999 999 999 999 998 416 016 061 235 2 × 2 = 1 + 0.999 999 999 999 999 996 832 032 122 470 4;
  • 46) 0.999 999 999 999 999 996 832 032 122 470 4 × 2 = 1 + 0.999 999 999 999 999 993 664 064 244 940 8;
  • 47) 0.999 999 999 999 999 993 664 064 244 940 8 × 2 = 1 + 0.999 999 999 999 999 987 328 128 489 881 6;

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).


4. 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:


0.000 000 060 536 876 844 707 876 443 862 825(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0000 1111 111(2)

5. Positive number before normalization:

0.000 000 060 536 876 844 707 876 443 862 825(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0000 1111 111(2)

6. 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:


0.000 000 060 536 876 844 707 876 443 862 825(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0000 1111 111(2) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0000 1111 111(2) × 20 =

1.0000 0100 0000 0000 1111 111(2) × 2-24


7. 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 0 (a positive number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.0000 0100 0000 0000 1111 111


8. 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)


9. 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;

10. 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)


11. 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. 000 0010 0000 0000 0111 1111 =

000 0010 0000 0000 0111 1111


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0111

Mantissa (23 bits) =
000 0010 0000 0000 0111 1111


Decimal number 0.000 000 060 536 876 844 707 876 443 862 825 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 000 0010 0000 0000 0111 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