-0.000 000 000 742 147 676 646 72 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 72(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 72(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 72| = 0.000 000 000 742 147 676 646 72


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 72.

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 72 × 2 = 0 + 0.000 000 001 484 295 353 293 44;
  • 2) 0.000 000 001 484 295 353 293 44 × 2 = 0 + 0.000 000 002 968 590 706 586 88;
  • 3) 0.000 000 002 968 590 706 586 88 × 2 = 0 + 0.000 000 005 937 181 413 173 76;
  • 4) 0.000 000 005 937 181 413 173 76 × 2 = 0 + 0.000 000 011 874 362 826 347 52;
  • 5) 0.000 000 011 874 362 826 347 52 × 2 = 0 + 0.000 000 023 748 725 652 695 04;
  • 6) 0.000 000 023 748 725 652 695 04 × 2 = 0 + 0.000 000 047 497 451 305 390 08;
  • 7) 0.000 000 047 497 451 305 390 08 × 2 = 0 + 0.000 000 094 994 902 610 780 16;
  • 8) 0.000 000 094 994 902 610 780 16 × 2 = 0 + 0.000 000 189 989 805 221 560 32;
  • 9) 0.000 000 189 989 805 221 560 32 × 2 = 0 + 0.000 000 379 979 610 443 120 64;
  • 10) 0.000 000 379 979 610 443 120 64 × 2 = 0 + 0.000 000 759 959 220 886 241 28;
  • 11) 0.000 000 759 959 220 886 241 28 × 2 = 0 + 0.000 001 519 918 441 772 482 56;
  • 12) 0.000 001 519 918 441 772 482 56 × 2 = 0 + 0.000 003 039 836 883 544 965 12;
  • 13) 0.000 003 039 836 883 544 965 12 × 2 = 0 + 0.000 006 079 673 767 089 930 24;
  • 14) 0.000 006 079 673 767 089 930 24 × 2 = 0 + 0.000 012 159 347 534 179 860 48;
  • 15) 0.000 012 159 347 534 179 860 48 × 2 = 0 + 0.000 024 318 695 068 359 720 96;
  • 16) 0.000 024 318 695 068 359 720 96 × 2 = 0 + 0.000 048 637 390 136 719 441 92;
  • 17) 0.000 048 637 390 136 719 441 92 × 2 = 0 + 0.000 097 274 780 273 438 883 84;
  • 18) 0.000 097 274 780 273 438 883 84 × 2 = 0 + 0.000 194 549 560 546 877 767 68;
  • 19) 0.000 194 549 560 546 877 767 68 × 2 = 0 + 0.000 389 099 121 093 755 535 36;
  • 20) 0.000 389 099 121 093 755 535 36 × 2 = 0 + 0.000 778 198 242 187 511 070 72;
  • 21) 0.000 778 198 242 187 511 070 72 × 2 = 0 + 0.001 556 396 484 375 022 141 44;
  • 22) 0.001 556 396 484 375 022 141 44 × 2 = 0 + 0.003 112 792 968 750 044 282 88;
  • 23) 0.003 112 792 968 750 044 282 88 × 2 = 0 + 0.006 225 585 937 500 088 565 76;
  • 24) 0.006 225 585 937 500 088 565 76 × 2 = 0 + 0.012 451 171 875 000 177 131 52;
  • 25) 0.012 451 171 875 000 177 131 52 × 2 = 0 + 0.024 902 343 750 000 354 263 04;
  • 26) 0.024 902 343 750 000 354 263 04 × 2 = 0 + 0.049 804 687 500 000 708 526 08;
  • 27) 0.049 804 687 500 000 708 526 08 × 2 = 0 + 0.099 609 375 000 001 417 052 16;
  • 28) 0.099 609 375 000 001 417 052 16 × 2 = 0 + 0.199 218 750 000 002 834 104 32;
  • 29) 0.199 218 750 000 002 834 104 32 × 2 = 0 + 0.398 437 500 000 005 668 208 64;
  • 30) 0.398 437 500 000 005 668 208 64 × 2 = 0 + 0.796 875 000 000 011 336 417 28;
  • 31) 0.796 875 000 000 011 336 417 28 × 2 = 1 + 0.593 750 000 000 022 672 834 56;
  • 32) 0.593 750 000 000 022 672 834 56 × 2 = 1 + 0.187 500 000 000 045 345 669 12;
  • 33) 0.187 500 000 000 045 345 669 12 × 2 = 0 + 0.375 000 000 000 090 691 338 24;
  • 34) 0.375 000 000 000 090 691 338 24 × 2 = 0 + 0.750 000 000 000 181 382 676 48;
  • 35) 0.750 000 000 000 181 382 676 48 × 2 = 1 + 0.500 000 000 000 362 765 352 96;
  • 36) 0.500 000 000 000 362 765 352 96 × 2 = 1 + 0.000 000 000 000 725 530 705 92;
  • 37) 0.000 000 000 000 725 530 705 92 × 2 = 0 + 0.000 000 000 001 451 061 411 84;
  • 38) 0.000 000 000 001 451 061 411 84 × 2 = 0 + 0.000 000 000 002 902 122 823 68;
  • 39) 0.000 000 000 002 902 122 823 68 × 2 = 0 + 0.000 000 000 005 804 245 647 36;
  • 40) 0.000 000 000 005 804 245 647 36 × 2 = 0 + 0.000 000 000 011 608 491 294 72;
  • 41) 0.000 000 000 011 608 491 294 72 × 2 = 0 + 0.000 000 000 023 216 982 589 44;
  • 42) 0.000 000 000 023 216 982 589 44 × 2 = 0 + 0.000 000 000 046 433 965 178 88;
  • 43) 0.000 000 000 046 433 965 178 88 × 2 = 0 + 0.000 000 000 092 867 930 357 76;
  • 44) 0.000 000 000 092 867 930 357 76 × 2 = 0 + 0.000 000 000 185 735 860 715 52;
  • 45) 0.000 000 000 185 735 860 715 52 × 2 = 0 + 0.000 000 000 371 471 721 431 04;
  • 46) 0.000 000 000 371 471 721 431 04 × 2 = 0 + 0.000 000 000 742 943 442 862 08;
  • 47) 0.000 000 000 742 943 442 862 08 × 2 = 0 + 0.000 000 001 485 886 885 724 16;
  • 48) 0.000 000 001 485 886 885 724 16 × 2 = 0 + 0.000 000 002 971 773 771 448 32;
  • 49) 0.000 000 002 971 773 771 448 32 × 2 = 0 + 0.000 000 005 943 547 542 896 64;
  • 50) 0.000 000 005 943 547 542 896 64 × 2 = 0 + 0.000 000 011 887 095 085 793 28;
  • 51) 0.000 000 011 887 095 085 793 28 × 2 = 0 + 0.000 000 023 774 190 171 586 56;
  • 52) 0.000 000 023 774 190 171 586 56 × 2 = 0 + 0.000 000 047 548 380 343 173 12;
  • 53) 0.000 000 047 548 380 343 173 12 × 2 = 0 + 0.000 000 095 096 760 686 346 24;
  • 54) 0.000 000 095 096 760 686 346 24 × 2 = 0 + 0.000 000 190 193 521 372 692 48;

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 72(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 72(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 72(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 72 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