-0.000 000 000 742 147 673 39 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 39(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
-0.000 000 000 742 147 673 39(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:

|-0.000 000 000 742 147 673 39| = 0.000 000 000 742 147 673 39


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: 0.000 000 000 742 147 673 39.

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) 0.000 000 000 742 147 673 39 × 2 = 0 + 0.000 000 001 484 295 346 78;
  • 2) 0.000 000 001 484 295 346 78 × 2 = 0 + 0.000 000 002 968 590 693 56;
  • 3) 0.000 000 002 968 590 693 56 × 2 = 0 + 0.000 000 005 937 181 387 12;
  • 4) 0.000 000 005 937 181 387 12 × 2 = 0 + 0.000 000 011 874 362 774 24;
  • 5) 0.000 000 011 874 362 774 24 × 2 = 0 + 0.000 000 023 748 725 548 48;
  • 6) 0.000 000 023 748 725 548 48 × 2 = 0 + 0.000 000 047 497 451 096 96;
  • 7) 0.000 000 047 497 451 096 96 × 2 = 0 + 0.000 000 094 994 902 193 92;
  • 8) 0.000 000 094 994 902 193 92 × 2 = 0 + 0.000 000 189 989 804 387 84;
  • 9) 0.000 000 189 989 804 387 84 × 2 = 0 + 0.000 000 379 979 608 775 68;
  • 10) 0.000 000 379 979 608 775 68 × 2 = 0 + 0.000 000 759 959 217 551 36;
  • 11) 0.000 000 759 959 217 551 36 × 2 = 0 + 0.000 001 519 918 435 102 72;
  • 12) 0.000 001 519 918 435 102 72 × 2 = 0 + 0.000 003 039 836 870 205 44;
  • 13) 0.000 003 039 836 870 205 44 × 2 = 0 + 0.000 006 079 673 740 410 88;
  • 14) 0.000 006 079 673 740 410 88 × 2 = 0 + 0.000 012 159 347 480 821 76;
  • 15) 0.000 012 159 347 480 821 76 × 2 = 0 + 0.000 024 318 694 961 643 52;
  • 16) 0.000 024 318 694 961 643 52 × 2 = 0 + 0.000 048 637 389 923 287 04;
  • 17) 0.000 048 637 389 923 287 04 × 2 = 0 + 0.000 097 274 779 846 574 08;
  • 18) 0.000 097 274 779 846 574 08 × 2 = 0 + 0.000 194 549 559 693 148 16;
  • 19) 0.000 194 549 559 693 148 16 × 2 = 0 + 0.000 389 099 119 386 296 32;
  • 20) 0.000 389 099 119 386 296 32 × 2 = 0 + 0.000 778 198 238 772 592 64;
  • 21) 0.000 778 198 238 772 592 64 × 2 = 0 + 0.001 556 396 477 545 185 28;
  • 22) 0.001 556 396 477 545 185 28 × 2 = 0 + 0.003 112 792 955 090 370 56;
  • 23) 0.003 112 792 955 090 370 56 × 2 = 0 + 0.006 225 585 910 180 741 12;
  • 24) 0.006 225 585 910 180 741 12 × 2 = 0 + 0.012 451 171 820 361 482 24;
  • 25) 0.012 451 171 820 361 482 24 × 2 = 0 + 0.024 902 343 640 722 964 48;
  • 26) 0.024 902 343 640 722 964 48 × 2 = 0 + 0.049 804 687 281 445 928 96;
  • 27) 0.049 804 687 281 445 928 96 × 2 = 0 + 0.099 609 374 562 891 857 92;
  • 28) 0.099 609 374 562 891 857 92 × 2 = 0 + 0.199 218 749 125 783 715 84;
  • 29) 0.199 218 749 125 783 715 84 × 2 = 0 + 0.398 437 498 251 567 431 68;
  • 30) 0.398 437 498 251 567 431 68 × 2 = 0 + 0.796 874 996 503 134 863 36;
  • 31) 0.796 874 996 503 134 863 36 × 2 = 1 + 0.593 749 993 006 269 726 72;
  • 32) 0.593 749 993 006 269 726 72 × 2 = 1 + 0.187 499 986 012 539 453 44;
  • 33) 0.187 499 986 012 539 453 44 × 2 = 0 + 0.374 999 972 025 078 906 88;
  • 34) 0.374 999 972 025 078 906 88 × 2 = 0 + 0.749 999 944 050 157 813 76;
  • 35) 0.749 999 944 050 157 813 76 × 2 = 1 + 0.499 999 888 100 315 627 52;
  • 36) 0.499 999 888 100 315 627 52 × 2 = 0 + 0.999 999 776 200 631 255 04;
  • 37) 0.999 999 776 200 631 255 04 × 2 = 1 + 0.999 999 552 401 262 510 08;
  • 38) 0.999 999 552 401 262 510 08 × 2 = 1 + 0.999 999 104 802 525 020 16;
  • 39) 0.999 999 104 802 525 020 16 × 2 = 1 + 0.999 998 209 605 050 040 32;
  • 40) 0.999 998 209 605 050 040 32 × 2 = 1 + 0.999 996 419 210 100 080 64;
  • 41) 0.999 996 419 210 100 080 64 × 2 = 1 + 0.999 992 838 420 200 161 28;
  • 42) 0.999 992 838 420 200 161 28 × 2 = 1 + 0.999 985 676 840 400 322 56;
  • 43) 0.999 985 676 840 400 322 56 × 2 = 1 + 0.999 971 353 680 800 645 12;
  • 44) 0.999 971 353 680 800 645 12 × 2 = 1 + 0.999 942 707 361 601 290 24;
  • 45) 0.999 942 707 361 601 290 24 × 2 = 1 + 0.999 885 414 723 202 580 48;
  • 46) 0.999 885 414 723 202 580 48 × 2 = 1 + 0.999 770 829 446 405 160 96;
  • 47) 0.999 770 829 446 405 160 96 × 2 = 1 + 0.999 541 658 892 810 321 92;
  • 48) 0.999 541 658 892 810 321 92 × 2 = 1 + 0.999 083 317 785 620 643 84;
  • 49) 0.999 083 317 785 620 643 84 × 2 = 1 + 0.998 166 635 571 241 287 68;
  • 50) 0.998 166 635 571 241 287 68 × 2 = 1 + 0.996 333 271 142 482 575 36;
  • 51) 0.996 333 271 142 482 575 36 × 2 = 1 + 0.992 666 542 284 965 150 72;
  • 52) 0.992 666 542 284 965 150 72 × 2 = 1 + 0.985 333 084 569 930 301 44;
  • 53) 0.985 333 084 569 930 301 44 × 2 = 1 + 0.970 666 169 139 860 602 88;
  • 54) 0.970 666 169 139 860 602 88 × 2 = 1 + 0.941 332 338 279 721 205 76;

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:


0.000 000 000 742 147 673 39(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 673 39(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:


0.000 000 000 742 147 673 39(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 -0.000 000 000 742 147 673 39 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