-1.570 796 326 794 896 619 231 311 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1.570 796 326 794 896 619 231 311(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
-1.570 796 326 794 896 619 231 311(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:

|-1.570 796 326 794 896 619 231 311| = 1.570 796 326 794 896 619 231 311


2. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

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.

1(10) =

1(2)


4. Convert to binary (base 2) the fractional part: 0.570 796 326 794 896 619 231 311.

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) 0.570 796 326 794 896 619 231 311 × 2 = 1 + 0.141 592 653 589 793 238 462 622;
  • 2) 0.141 592 653 589 793 238 462 622 × 2 = 0 + 0.283 185 307 179 586 476 925 244;
  • 3) 0.283 185 307 179 586 476 925 244 × 2 = 0 + 0.566 370 614 359 172 953 850 488;
  • 4) 0.566 370 614 359 172 953 850 488 × 2 = 1 + 0.132 741 228 718 345 907 700 976;
  • 5) 0.132 741 228 718 345 907 700 976 × 2 = 0 + 0.265 482 457 436 691 815 401 952;
  • 6) 0.265 482 457 436 691 815 401 952 × 2 = 0 + 0.530 964 914 873 383 630 803 904;
  • 7) 0.530 964 914 873 383 630 803 904 × 2 = 1 + 0.061 929 829 746 767 261 607 808;
  • 8) 0.061 929 829 746 767 261 607 808 × 2 = 0 + 0.123 859 659 493 534 523 215 616;
  • 9) 0.123 859 659 493 534 523 215 616 × 2 = 0 + 0.247 719 318 987 069 046 431 232;
  • 10) 0.247 719 318 987 069 046 431 232 × 2 = 0 + 0.495 438 637 974 138 092 862 464;
  • 11) 0.495 438 637 974 138 092 862 464 × 2 = 0 + 0.990 877 275 948 276 185 724 928;
  • 12) 0.990 877 275 948 276 185 724 928 × 2 = 1 + 0.981 754 551 896 552 371 449 856;
  • 13) 0.981 754 551 896 552 371 449 856 × 2 = 1 + 0.963 509 103 793 104 742 899 712;
  • 14) 0.963 509 103 793 104 742 899 712 × 2 = 1 + 0.927 018 207 586 209 485 799 424;
  • 15) 0.927 018 207 586 209 485 799 424 × 2 = 1 + 0.854 036 415 172 418 971 598 848;
  • 16) 0.854 036 415 172 418 971 598 848 × 2 = 1 + 0.708 072 830 344 837 943 197 696;
  • 17) 0.708 072 830 344 837 943 197 696 × 2 = 1 + 0.416 145 660 689 675 886 395 392;
  • 18) 0.416 145 660 689 675 886 395 392 × 2 = 0 + 0.832 291 321 379 351 772 790 784;
  • 19) 0.832 291 321 379 351 772 790 784 × 2 = 1 + 0.664 582 642 758 703 545 581 568;
  • 20) 0.664 582 642 758 703 545 581 568 × 2 = 1 + 0.329 165 285 517 407 091 163 136;
  • 21) 0.329 165 285 517 407 091 163 136 × 2 = 0 + 0.658 330 571 034 814 182 326 272;
  • 22) 0.658 330 571 034 814 182 326 272 × 2 = 1 + 0.316 661 142 069 628 364 652 544;
  • 23) 0.316 661 142 069 628 364 652 544 × 2 = 0 + 0.633 322 284 139 256 729 305 088;
  • 24) 0.633 322 284 139 256 729 305 088 × 2 = 1 + 0.266 644 568 278 513 458 610 176;

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:


0.570 796 326 794 896 619 231 311(10) =

0.1001 0010 0001 1111 1011 0101(2)

6. Positive number before normalization:

1.570 796 326 794 896 619 231 311(10) =

1.1001 0010 0001 1111 1011 0101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.570 796 326 794 896 619 231 311(10) =

1.1001 0010 0001 1111 1011 0101(2) =

1.1001 0010 0001 1111 1011 0101(2) × 20


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): 0

Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(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;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

127(10) =

0111 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 100 1001 0000 1111 1101 1010 1 =

100 1001 0000 1111 1101 1010


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) =
0111 1111

Mantissa (23 bits) =
100 1001 0000 1111 1101 1010


Decimal number -1.570 796 326 794 896 619 231 311 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 1111 - 100 1001 0000 1111 1101 1010

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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