-0.000 000 000 742 147 676 646 709 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 709 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
-0.000 000 000 742 147 676 646 709 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:

|-0.000 000 000 742 147 676 646 709 5| = 0.000 000 000 742 147 676 646 709 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: 0.000 000 000 742 147 676 646 709 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) 0.000 000 000 742 147 676 646 709 5 × 2 = 0 + 0.000 000 001 484 295 353 293 419;
  • 2) 0.000 000 001 484 295 353 293 419 × 2 = 0 + 0.000 000 002 968 590 706 586 838;
  • 3) 0.000 000 002 968 590 706 586 838 × 2 = 0 + 0.000 000 005 937 181 413 173 676;
  • 4) 0.000 000 005 937 181 413 173 676 × 2 = 0 + 0.000 000 011 874 362 826 347 352;
  • 5) 0.000 000 011 874 362 826 347 352 × 2 = 0 + 0.000 000 023 748 725 652 694 704;
  • 6) 0.000 000 023 748 725 652 694 704 × 2 = 0 + 0.000 000 047 497 451 305 389 408;
  • 7) 0.000 000 047 497 451 305 389 408 × 2 = 0 + 0.000 000 094 994 902 610 778 816;
  • 8) 0.000 000 094 994 902 610 778 816 × 2 = 0 + 0.000 000 189 989 805 221 557 632;
  • 9) 0.000 000 189 989 805 221 557 632 × 2 = 0 + 0.000 000 379 979 610 443 115 264;
  • 10) 0.000 000 379 979 610 443 115 264 × 2 = 0 + 0.000 000 759 959 220 886 230 528;
  • 11) 0.000 000 759 959 220 886 230 528 × 2 = 0 + 0.000 001 519 918 441 772 461 056;
  • 12) 0.000 001 519 918 441 772 461 056 × 2 = 0 + 0.000 003 039 836 883 544 922 112;
  • 13) 0.000 003 039 836 883 544 922 112 × 2 = 0 + 0.000 006 079 673 767 089 844 224;
  • 14) 0.000 006 079 673 767 089 844 224 × 2 = 0 + 0.000 012 159 347 534 179 688 448;
  • 15) 0.000 012 159 347 534 179 688 448 × 2 = 0 + 0.000 024 318 695 068 359 376 896;
  • 16) 0.000 024 318 695 068 359 376 896 × 2 = 0 + 0.000 048 637 390 136 718 753 792;
  • 17) 0.000 048 637 390 136 718 753 792 × 2 = 0 + 0.000 097 274 780 273 437 507 584;
  • 18) 0.000 097 274 780 273 437 507 584 × 2 = 0 + 0.000 194 549 560 546 875 015 168;
  • 19) 0.000 194 549 560 546 875 015 168 × 2 = 0 + 0.000 389 099 121 093 750 030 336;
  • 20) 0.000 389 099 121 093 750 030 336 × 2 = 0 + 0.000 778 198 242 187 500 060 672;
  • 21) 0.000 778 198 242 187 500 060 672 × 2 = 0 + 0.001 556 396 484 375 000 121 344;
  • 22) 0.001 556 396 484 375 000 121 344 × 2 = 0 + 0.003 112 792 968 750 000 242 688;
  • 23) 0.003 112 792 968 750 000 242 688 × 2 = 0 + 0.006 225 585 937 500 000 485 376;
  • 24) 0.006 225 585 937 500 000 485 376 × 2 = 0 + 0.012 451 171 875 000 000 970 752;
  • 25) 0.012 451 171 875 000 000 970 752 × 2 = 0 + 0.024 902 343 750 000 001 941 504;
  • 26) 0.024 902 343 750 000 001 941 504 × 2 = 0 + 0.049 804 687 500 000 003 883 008;
  • 27) 0.049 804 687 500 000 003 883 008 × 2 = 0 + 0.099 609 375 000 000 007 766 016;
  • 28) 0.099 609 375 000 000 007 766 016 × 2 = 0 + 0.199 218 750 000 000 015 532 032;
  • 29) 0.199 218 750 000 000 015 532 032 × 2 = 0 + 0.398 437 500 000 000 031 064 064;
  • 30) 0.398 437 500 000 000 031 064 064 × 2 = 0 + 0.796 875 000 000 000 062 128 128;
  • 31) 0.796 875 000 000 000 062 128 128 × 2 = 1 + 0.593 750 000 000 000 124 256 256;
  • 32) 0.593 750 000 000 000 124 256 256 × 2 = 1 + 0.187 500 000 000 000 248 512 512;
  • 33) 0.187 500 000 000 000 248 512 512 × 2 = 0 + 0.375 000 000 000 000 497 025 024;
  • 34) 0.375 000 000 000 000 497 025 024 × 2 = 0 + 0.750 000 000 000 000 994 050 048;
  • 35) 0.750 000 000 000 000 994 050 048 × 2 = 1 + 0.500 000 000 000 001 988 100 096;
  • 36) 0.500 000 000 000 001 988 100 096 × 2 = 1 + 0.000 000 000 000 003 976 200 192;
  • 37) 0.000 000 000 000 003 976 200 192 × 2 = 0 + 0.000 000 000 000 007 952 400 384;
  • 38) 0.000 000 000 000 007 952 400 384 × 2 = 0 + 0.000 000 000 000 015 904 800 768;
  • 39) 0.000 000 000 000 015 904 800 768 × 2 = 0 + 0.000 000 000 000 031 809 601 536;
  • 40) 0.000 000 000 000 031 809 601 536 × 2 = 0 + 0.000 000 000 000 063 619 203 072;
  • 41) 0.000 000 000 000 063 619 203 072 × 2 = 0 + 0.000 000 000 000 127 238 406 144;
  • 42) 0.000 000 000 000 127 238 406 144 × 2 = 0 + 0.000 000 000 000 254 476 812 288;
  • 43) 0.000 000 000 000 254 476 812 288 × 2 = 0 + 0.000 000 000 000 508 953 624 576;
  • 44) 0.000 000 000 000 508 953 624 576 × 2 = 0 + 0.000 000 000 001 017 907 249 152;
  • 45) 0.000 000 000 001 017 907 249 152 × 2 = 0 + 0.000 000 000 002 035 814 498 304;
  • 46) 0.000 000 000 002 035 814 498 304 × 2 = 0 + 0.000 000 000 004 071 628 996 608;
  • 47) 0.000 000 000 004 071 628 996 608 × 2 = 0 + 0.000 000 000 008 143 257 993 216;
  • 48) 0.000 000 000 008 143 257 993 216 × 2 = 0 + 0.000 000 000 016 286 515 986 432;
  • 49) 0.000 000 000 016 286 515 986 432 × 2 = 0 + 0.000 000 000 032 573 031 972 864;
  • 50) 0.000 000 000 032 573 031 972 864 × 2 = 0 + 0.000 000 000 065 146 063 945 728;
  • 51) 0.000 000 000 065 146 063 945 728 × 2 = 0 + 0.000 000 000 130 292 127 891 456;
  • 52) 0.000 000 000 130 292 127 891 456 × 2 = 0 + 0.000 000 000 260 584 255 782 912;
  • 53) 0.000 000 000 260 584 255 782 912 × 2 = 0 + 0.000 000 000 521 168 511 565 824;
  • 54) 0.000 000 000 521 168 511 565 824 × 2 = 0 + 0.000 000 001 042 337 023 131 648;

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:


0.000 000 000 742 147 676 646 709 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 709 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 709 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 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 0000

Mantissa (23 bits) =
100 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 709 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 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