-0.000 000 000 742 147 676 575 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 575(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
-0.000 000 000 742 147 676 575(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:

|-0.000 000 000 742 147 676 575| = 0.000 000 000 742 147 676 575


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: 0.000 000 000 742 147 676 575.

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) 0.000 000 000 742 147 676 575 × 2 = 0 + 0.000 000 001 484 295 353 15;
  • 2) 0.000 000 001 484 295 353 15 × 2 = 0 + 0.000 000 002 968 590 706 3;
  • 3) 0.000 000 002 968 590 706 3 × 2 = 0 + 0.000 000 005 937 181 412 6;
  • 4) 0.000 000 005 937 181 412 6 × 2 = 0 + 0.000 000 011 874 362 825 2;
  • 5) 0.000 000 011 874 362 825 2 × 2 = 0 + 0.000 000 023 748 725 650 4;
  • 6) 0.000 000 023 748 725 650 4 × 2 = 0 + 0.000 000 047 497 451 300 8;
  • 7) 0.000 000 047 497 451 300 8 × 2 = 0 + 0.000 000 094 994 902 601 6;
  • 8) 0.000 000 094 994 902 601 6 × 2 = 0 + 0.000 000 189 989 805 203 2;
  • 9) 0.000 000 189 989 805 203 2 × 2 = 0 + 0.000 000 379 979 610 406 4;
  • 10) 0.000 000 379 979 610 406 4 × 2 = 0 + 0.000 000 759 959 220 812 8;
  • 11) 0.000 000 759 959 220 812 8 × 2 = 0 + 0.000 001 519 918 441 625 6;
  • 12) 0.000 001 519 918 441 625 6 × 2 = 0 + 0.000 003 039 836 883 251 2;
  • 13) 0.000 003 039 836 883 251 2 × 2 = 0 + 0.000 006 079 673 766 502 4;
  • 14) 0.000 006 079 673 766 502 4 × 2 = 0 + 0.000 012 159 347 533 004 8;
  • 15) 0.000 012 159 347 533 004 8 × 2 = 0 + 0.000 024 318 695 066 009 6;
  • 16) 0.000 024 318 695 066 009 6 × 2 = 0 + 0.000 048 637 390 132 019 2;
  • 17) 0.000 048 637 390 132 019 2 × 2 = 0 + 0.000 097 274 780 264 038 4;
  • 18) 0.000 097 274 780 264 038 4 × 2 = 0 + 0.000 194 549 560 528 076 8;
  • 19) 0.000 194 549 560 528 076 8 × 2 = 0 + 0.000 389 099 121 056 153 6;
  • 20) 0.000 389 099 121 056 153 6 × 2 = 0 + 0.000 778 198 242 112 307 2;
  • 21) 0.000 778 198 242 112 307 2 × 2 = 0 + 0.001 556 396 484 224 614 4;
  • 22) 0.001 556 396 484 224 614 4 × 2 = 0 + 0.003 112 792 968 449 228 8;
  • 23) 0.003 112 792 968 449 228 8 × 2 = 0 + 0.006 225 585 936 898 457 6;
  • 24) 0.006 225 585 936 898 457 6 × 2 = 0 + 0.012 451 171 873 796 915 2;
  • 25) 0.012 451 171 873 796 915 2 × 2 = 0 + 0.024 902 343 747 593 830 4;
  • 26) 0.024 902 343 747 593 830 4 × 2 = 0 + 0.049 804 687 495 187 660 8;
  • 27) 0.049 804 687 495 187 660 8 × 2 = 0 + 0.099 609 374 990 375 321 6;
  • 28) 0.099 609 374 990 375 321 6 × 2 = 0 + 0.199 218 749 980 750 643 2;
  • 29) 0.199 218 749 980 750 643 2 × 2 = 0 + 0.398 437 499 961 501 286 4;
  • 30) 0.398 437 499 961 501 286 4 × 2 = 0 + 0.796 874 999 923 002 572 8;
  • 31) 0.796 874 999 923 002 572 8 × 2 = 1 + 0.593 749 999 846 005 145 6;
  • 32) 0.593 749 999 846 005 145 6 × 2 = 1 + 0.187 499 999 692 010 291 2;
  • 33) 0.187 499 999 692 010 291 2 × 2 = 0 + 0.374 999 999 384 020 582 4;
  • 34) 0.374 999 999 384 020 582 4 × 2 = 0 + 0.749 999 998 768 041 164 8;
  • 35) 0.749 999 998 768 041 164 8 × 2 = 1 + 0.499 999 997 536 082 329 6;
  • 36) 0.499 999 997 536 082 329 6 × 2 = 0 + 0.999 999 995 072 164 659 2;
  • 37) 0.999 999 995 072 164 659 2 × 2 = 1 + 0.999 999 990 144 329 318 4;
  • 38) 0.999 999 990 144 329 318 4 × 2 = 1 + 0.999 999 980 288 658 636 8;
  • 39) 0.999 999 980 288 658 636 8 × 2 = 1 + 0.999 999 960 577 317 273 6;
  • 40) 0.999 999 960 577 317 273 6 × 2 = 1 + 0.999 999 921 154 634 547 2;
  • 41) 0.999 999 921 154 634 547 2 × 2 = 1 + 0.999 999 842 309 269 094 4;
  • 42) 0.999 999 842 309 269 094 4 × 2 = 1 + 0.999 999 684 618 538 188 8;
  • 43) 0.999 999 684 618 538 188 8 × 2 = 1 + 0.999 999 369 237 076 377 6;
  • 44) 0.999 999 369 237 076 377 6 × 2 = 1 + 0.999 998 738 474 152 755 2;
  • 45) 0.999 998 738 474 152 755 2 × 2 = 1 + 0.999 997 476 948 305 510 4;
  • 46) 0.999 997 476 948 305 510 4 × 2 = 1 + 0.999 994 953 896 611 020 8;
  • 47) 0.999 994 953 896 611 020 8 × 2 = 1 + 0.999 989 907 793 222 041 6;
  • 48) 0.999 989 907 793 222 041 6 × 2 = 1 + 0.999 979 815 586 444 083 2;
  • 49) 0.999 979 815 586 444 083 2 × 2 = 1 + 0.999 959 631 172 888 166 4;
  • 50) 0.999 959 631 172 888 166 4 × 2 = 1 + 0.999 919 262 345 776 332 8;
  • 51) 0.999 919 262 345 776 332 8 × 2 = 1 + 0.999 838 524 691 552 665 6;
  • 52) 0.999 838 524 691 552 665 6 × 2 = 1 + 0.999 677 049 383 105 331 2;
  • 53) 0.999 677 049 383 105 331 2 × 2 = 1 + 0.999 354 098 766 210 662 4;
  • 54) 0.999 354 098 766 210 662 4 × 2 = 1 + 0.998 708 197 532 421 324 8;

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:


0.000 000 000 742 147 676 575(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 575(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:


0.000 000 000 742 147 676 575(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 -0.000 000 000 742 147 676 575 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